Talk:Horsepower

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I know that the horsepower ratings for vintage cars is very different from the modern BHP rating. For example the 9hp 2CV engine was not a tenth of the power of a modern small car, but probably about a third.

Does anyone know how these horespowers relate to BHP?

-- Chris Q 12:32 Sep 17, 2002 (UTC)

See tax horsepower -- Egil 23:29 Apr 24, 2003 (UTC)

[edit] Now

The article dosen't explain how horsepower is measured today. [This text was submitted unsigned by User:Commander Ball]

Do you mean the machinery involved, or the way it is calculated? Most of this is not actually done in horsepower anymore, what with the worldwide adoption of the metric system; thus it is better discussed at a system-independent place. —Morven 06:14, Jul 9, 2004 (UTC)

[edit] Real power of the horse?

I read somewhere that HP is defined smaller than real power of the horse is, in order to make better marketing effect for first engines. Can someone confirm this, plz? saigon_from_europe

[edit] Original horse power

The original horsepower unit was selected to show potential buyers of the new steam engines how much they would save by switching from horses to steam. The horsepower unit represented the rate of work a horse could do on a continous basis (all day long). Thats why its only a small fraction of the maximum capability of a horse.

[edit] Derivation for HP Equation

Could someone possibly go through the derivation for equating Horsepower to RPM and Torque of an engine or motor. I know that $\mathbf{HP} = {\mathbf{Torque} * \mathbf{RPM} \over 5252}$ is the equation, I just seem to be at a loss as to show the derivation as my algebra skills aren't exactly top notch. Jason Gillman Jr.

First of all, that won't give you the relationship between typical horsepower ratings and typical torque ratings, because each is usually expressed at a different RPM, and neither of the associated curves are linear functions of the rotational speed.

What units are you using for torque? That's part of it. For the particular numbers you have, you need to use lbf·ft (or ft·lbf if you prefer).

Express horsepower in terms of feet, pounds-force, and some time unit. The article does that for you.

A revolution is 2×&pi radians. Convert revolutions per minute to radians per minute or per second, whchever time unit you used with the horsepower. Radians are actually dimensionless. Ignore them in your final units.

Multiply 5252 by π and round to 4 significant digits, since that's all you have in that 5252 number. Then double it (or, you could just multiply by 2π in the first place). Is the familiar number a helpful hint? Gene Nygaard 01:11, 6 Apr 2005 (UTC)

I wrote this up at torque. Samw 03:20, 9 Apr 2005 (UTC)

[edit] Pound mass

The article said "15,000 pound mass" - this is clearly wrong.This number is repeated in the SI approximation. I'll change this. I always thought that the "pound-foot" system unit of mass is the slug (pounds weight divided by G). Wikipedia says there's a pound mass and a pound force but I'm confused...I thought a pound was defined as the *weight* of the mass, not the mass? --Wtshymanski 03:42, 7 Apr 2005 (UTC)

Like you said, you are confused. THe unfortunate thing is that there are far too many teachers in the same boat. Maybe a look at poundal and pound-force will help you out. Of course, you could actually lookup the definition of a pound, too—here's the official U.S. definition, and as it points out, the official definition in the rest of the world as well.[1]

It is, in fact, the pound-force which is the recent bastardization, a unit never well-defined before the 20th century, and one which even today doesn't have an official definition. -- Gene Nygaard 04:43, 7 Apr 2005 (UTC)

I forgot to mention that you were not so confused as to not catch that 15,000 pound error. Good job on that. Gene Nygaard 04:45, 7 Apr 2005 (UTC)

Luckily, I never have to worry about pounds (forece) and pounds (mass)- I understand kilograms and newtons adequately. But I knew the "15,000" didn't fit! --Wtshymanski 03:25, 8 Apr 2005 (UTC)

[edit] Determining horsepower

Is there a simple method that a common layman could use to determine approximate horsepower of a car's engine?? Jaberwocky6669 18:37, May 3, 2005 (UTC)

Unfortunately, no. A quick look at many engines will show you that engine size is not necessarily all that indicative of horsepower (all else being equal, of course, it often will be, but all else isn't equal) —Morven 19:45, May 3, 2005 (UTC)

Sure there is: take the kilowatt value, multiply by four and divide by three. Christoph Päper 13:49, 4 May 2005 (UTC)

Very funny ;-)

There are a couple of ways you can compute the average power at the wheels. The engine power is slightly greater due to drive train losses.

1. Measure acceleration of the car (e.g. 0-60 time) and calculate the power based on vehicle mass

2. Climb a steep hill and use the climb rate and vehicle mass to compute power for a given throttle setting

The first method is probably easier (and safer). I'll post the exact equations if I have time. Madhu 18:59, 5 March 2007 (UTC)

[edit] Horse-power machine

A horse-power is also a device used to power some other machine, where a horse or several horses were walked in a circle or on a belt to provide power. An example in a historical text is here: http://bchs.kearney.net/BTales_199303.html

Unless I miss my guess, it is the name of this machine from which the horsepower unit took its name, since this is what steam engines were devised to replace.

There were many machines that were operated by horses. The cited URL uses "horse power" as the adjective for how the machine is operated and not for the machine itself. I'm not aware of any machine named "horse power". But yes, clearly "horsepower" comes from the way machines were powered with horses. Samw 03:26, 24 May 2005 (UTC)

It is my impression that there were modular "horse power" machines that could be used to power many other machines, such as threshers, water pumps, etc., much as early belt-drive stationary steam engines could be attached to different machines to power them. I have seen a horse power machine on display at the Pioneer Auto exhibit in Murdo, South Dakota.

Interesting. I guess it makes sense to re-use portions of the machine that are on the "horse-end": the harnesses, the main rotor, etc. If you have references, it would make an interesting historical footnote on alternate historical uses of the word "horsepower". Samw 03:26, 29 May 2005 (UTC)

OK, here are some references. In the Papers of John C. Calhoun, 20 September 1845 entry, a horse power machine is mentioned several times. This work has been edited by Clyde N. Wilson and published in book form in 1995 by the University of South Carolina Press. The salient text is available on Google Print[2]

Additionally, a historical advertising postcard advocates "A. W. Gray's Son's Manufactures of Horse Powers." An image is available here: http://www.edhines.com/19591,%20ILLUSTRATED,%20MIDDLETOWN%20SPRINGS%20VERMONT%20HORSE%20POWER%20AD%20WITH%20STAMP%20COLLAR%20%2095.00.jpg

Another source is the caption to a photograph in the collection of The Historical Museum of Bonnyville in Canada. This caption reads "Machine called « Horse Power » supplying power for threshing". Although the website where I saw this picture presents the caption as html text, and not as part of the historcial image, I'm assuming that the caption is based on expert historical knowledge. The image is avaliable here: http://collections.ic.gc.ca/bonnyville/english/popup_photo/colonisation5.html

Great references, thanks! All take a shot at adding this to the article but you're welcome to do this yourself. Samw 05:20, 5 November 2005 (UTC)

I started a new article for this and linked to it at the top of this article. Hope that's cool.

Horse power (machine) looks great! Samw 04:23, 8 November 2005 (UTC)

[edit] Morven's comments from User talk:Bobblewik

Horsepower is a complicated issue. Part of it is to do with there being two definitions of horsepower: Imperial horsepower (1 hp = 33,000 ft·lbf·min^-1) and German pferdstarke (nowadays defined in terms of SI units: 1 PS = 75 kp·m/s = 735.49875 W). The two thus differ slightly; by my calculations, 1 hp = 1.014 PS.

Fairly often, in non US markets, a power rating is given in 'hp' in English-language publication, but the values are actually 'PS' unconverted. This is even sometimes true for British or American products sold on the world market.

In the US market for automobiles, true hp is used.

The second problem is how the power is measured. In the modern day, power measurements are in SAE net horsepower in the US, and DIN net (kW|horsepower) or JIS net (kW|horsepower) in most other markets. The testing method for both is the same; horsepower is measured at the engine flywheel with all engine accessories attached, alternator fully loaded, emissions-control equipment installed, and the production exhaust system attached. Thus, the difference between SAE and DIN/JIS ratings is largely explainable by the hp/PS difference above.

Before 1972, US manufacturers used SAE gross horsepower. This tested the engine's power at the flywheel without most accessories attached, no pollution control equipment, and an open, unrestricted exhaust system. This gave power figures approx 20% higher than SAE net, although there is much variation. Thus power figures for 1972 and later US cars cannot be directly compared with those quoted for earlier vehicles. A failure to understand this change exacerbated the impression left on the US consumer that emissions and fuel-economy laws in the 1970s ruined US cars. There was, indeed, much of a reduction in power, but the change in measurement method exaggerated it.

Added to all this, of course, is the fact that horsepower figures in the press are advertising numbers rather than engineering numbers, and thus are often rounded roughly, exaggerated or even minimised. It has in fact been quite common for mass-market manufacturers to understate the power of their high-performance vehicles, perhaps to avoid bad publicity or high insurance charges for the new owners. This affects figures in both kW and hp.

I would suggest that when we quote horsepower figures in car articles, we should state which hp is being talked about. For US-market automobiles in the modern day, we should state hp (SAE net) or SAE net horsepower, I think. It is safe to perform conversions to kW using the standard ratio, since SAE hp is measured in the same manner as DIN kW. Older US-market automobiles should be quoted as hp (SAE gross). We should also flag kW conversions for those vehicles as (gross) as a warning that they are not directly comparable to DIN kW figures.

In general, we should obtain the original kW figures for vehicles sold outside the US, since in most modern cases this will be the original measurement and the hp a conversion. We should also flag hp conversions as to whether they are DIN or JIS net hp (which are generally actually PS) or SAE net.

I am not sure which figures are generally used in the UK market; the figures are normally quoted in 'brake horsepower' (bhp), but what this exactly means, I'm unsure. I suspect it is probably quite equivalent to SAE net, but more research is probably needed here.

I hope this complicated explanation helps more than it hinders! —Morven 13:34, May 26, 2005 (UTC)

I copied the above from User talk:Bobblewik to make sure the information don't get lost. Some of it is here already, but much isn't. Unfortunately it is not a topic I know well enough to integrate it in. -- ALoan (Talk) 12:26, 27 May 2005 (UTC)

On the basis of all the help from everybody, I have been correcting the errors in Mercedes-Benz articles. I note that Morven said:

a power rating is given in 'hp' in English-language publication, but the values are actually 'PS' unconverted. This is even sometimes true for British or American products sold on the world market

I see now that it can even be inconsistent for the same product, language and region. For example, Mercedes Benz USA quotes the 150 kW engine as having 201 hp (presumably SAE net) in one part of the site and as having 204 hp (presumably DIN) in another part of the site. Sigh. Bobblewik (talk) 18:39, 28 May 2005 (UTC)

There you are, going off half-cocked again, Bobblewik. Please do not go around changing these numbers to "hp (DIN)". Using the symbol "hp" for the metric horsepower is unacceptable; spell out metric horsepower or one of its other names, or use a symbol which is acceptable. The symbol PS is not only the German symbol for this unit, but also the one most often used in English, though other symbols such as CV are occasionally seen. Then, if you are indeed sure that this is the original measurement, please list the original measurement first, followed by the conversion to kilowatts. If you don't think it is the original measurement, don't list it; list "hp" (English horsepower) instead. And please, go back and fix the articles you have messed up now. Gene Nygaard 19:00, May 28, 2005 (UTC)

I would strongly recommend against ever using "CV" to mean anything except French tax horsepower, unrelated to real power ratings. The general English-language exposure to the term is purely for that purpose.

For European and Japanese vehicles, I would suggest never using an offically listed "hp" rating unless we are absolutely sure that it was the original measurement. In almost all these cases, the original measurement would be the kW or MW one. We should then provide a hp conversion in terms of Imperial/US customary horsepower, using the standard conversion ratio. Except in cases where we are sure that the original design / measurement was in PS ("metric horsepower"), we should not use that unit in Wikipedia. All it adds is confusion to the reader.

For US-market cars, we should differentiate between SAE gross and SAE net horsepower. At the very least, all ratings in gross horsepower should be marked as such, to indicate that they cannot be compared with modern power ratings (net power being the accepted modern standard worldwide). —Morven 07:21, May 29, 2005 (UTC)

Gene Nygaard has told me that he is unhappy with my contributions. So I am asking for others to try to help fix the articles. I raised this issue of unreliable horsepower values and have added kilowatt values that my research makes me believe are accurate. In the absence of guidance, I changed the ambiguous term 'hp' to the term 'hp DIN'. If I should have changed 'hp' to 'PS' or 'CV', then mea culpa (my mistake). I have my flaws as do we all, but I have acted in good faith. I don't like being the target of expressions like 'half-cocked' and 'screwups'. Gene, we are both champions of good use of units, I respect your knowledge and dedication to research about units. Please treat me as your friend or at least collaborator, I am certainly not your enemy.

I ask Gene and others to examine the edits that I have made to car articles and try to improve them. I also ask for community guidance on terminology. I am prepared to put in some effort to fix the problems caused by the use of 'hp' as a primary value. I welcome positive actions and positive thoughts in support of this. Bobblewik (talk) 12:19, 29 May 2005 (UTC)

Sorry to be so abrupt about it, just commenting on your habits of rushing off to do a huge number of changes before being sure that you've gotten it right. Did you even notice that User:Morven used "PS" to distinguish from "hp"? I'm glad to see that you are willing to put in some effort into tracking down the changes you've already made; from what I've seen before, that's something you rarely do, which is why I made sure to categorize your changes this time as a "screw-up".

The fact of the matter is that most of these measurements (probably even for current production) were made in either English horsepower or metric horsepower, with the kilowatts figure being a conversion, and almost all for the majority of the articles which include measurements made at some time in the past. I urge you and all contributors to be aware of this problem, and if the source of your information lists numbers in both horsepower and kilowatts, be sure to check to see which horsepower was used there before putting it into the article, and disambiguate them here even if your source didn't do so. As I've said before, I think the best way to do that is to use "PS" (linked to horsepower on first use) as the symbol for metric horsepower, though if some other method is already used, I wouldn't bother changing it unless it uses "hp" (even with your parenthetical DIN this seems unclear). It's not only that "hp (DIN)" has never been explained in this "horsepower" article itself, but in fact the "DIN" is never mentioned at all in the article itself, and you did not provide either a spelled out version of DIN or a link to its article DIN either. Gene Nygaard 12:44, May 29, 2005 (UTC)

After some research I have discovered the there is a European Directive on this: 80/1269/EEC (and modified by: 88/195/EEC, 89/491/EEC, 97/21/EC and 1999/99/EC).

This Directive harmonises national legislation on the method of measuring engine power that must be used to indicate the engine power of a vehicle type.

Henceforth, no Member State may refuse to grant EEC type approval or national type approval in respect of a vehicle, or refuse or prohibit the sale, registration, entry into service or use of a vehicle on grounds relating to its engine power, if this has been determined in accordance with the Annexes to the Directive.

So perhaps the method of measurement itself has been harmonised in Europe, or will harmonise. I hope that somebody can make something of that. Bobblewik (talk) 01:14, 30 May 2005 (UTC)

So let's cut this down to the convention for Wikipedia. We already discussed the issue at Wikipedia:WikiProject Automobiles/Conventions and decided to use hp (presumably SAE net or gross depending on time period) or kW depending on market for this unit. I am opposed to using PS or metric horsepower - it's just plain confusing. One exception is if the metric hp number is "significant" - as in the Bugatti Veyron with it's 1001 hp (metric) engine. I will propose an ammendment over there to always specify "hp SAE" rather than just "hp" and see what people think. Or we could just link the first occurrence of "hp" to horsepower#hp (SAE) and let people use their intiution that all remaining references to hp are for SAE numbers. I like this better. We DO have to be precise, and I don't want to go around using random units everywhere.

So let's decide here and now:

I propose never using metric horsepower as a primary unit unless it's a special case like the Bugatti. And then we should be very specific in including the SAE number.
We've already decided to always include both hp and kW, and that's all well and good.
As for notation, if we are to specify SAE, I would do it without parentheses since we are using the parens for kW already. So we say "100 hp SAE (74 kW)". I'd rather not include the SAE at all, but this is a reasonable alternative.

--SFoskett 14:39, May 31, 2005 (UTC)

As I've said in Wikipedia Talk:WikiProject Automobiles/Conventions, converting PS values to hp would be even more confusing.

As an example, let's look at the MG ZR 160. A british car, should use bhp, right? Wrong. The 160 bhp are actually the PS value, 118 kW, which would be 158 hp, but even in the UK, what is actually PS gets called bhp. The number "158" isn't used anywhere, and to use it would prompt any newbie user to immediately revert to what he would perceive as the real value.

Now, let's take a look at the Dodge Viper SRT-10. Advertised as 500 hp (373 kW). In Europe, this would be 507 PS, but Chrysler actually keeps the original 500 hp value here, without correcting the kW (which, in Europe, would be 368).

Next, we have the Mercedes C 320. 160 kW, which are 218 PS in Europe and 215 hp in the USA.

Finally, let's see the Volkswagen Golf GTI (Mk IV). Advertised as having 180 hp, both in Europe and the USA. Strangely, the VW USA website tells us the car has 132 kW, which is 180 PS, or 177 hp.

So, my proposal is to retain the original PS values for all European and Japanese models (except for engines/models available exclusively in North America, like the 2.4L Mitsubishi Lancer), hp values for all American models (except for engines/models available everywhere else, such as 1.6L Neon and the Diesel-powered PT Cruiser) and use the kW to indicate the engine's true power. --Pc13 16:42, 2005 May 31 (UTC)

[edit] Coping with ambiguity in the term 'horsepower'

Discussion moved from Bobblewik talk but added to ongoing discussion

Will you please go back and fix all the articles you have screwed up by using they symbol "hp" for metric horsepower? See Talk:Horsepower. If you have actually gone to the trouble of disambiguating careless use of horsepower, make it clear by not using the symbol for the English units. If those are the original units, either spell out metric horsepower, or use PS for them, with that link to horsepower on first use&the metric horsepower units are discussed there. That symbol from the abbreviation of the German name is the most common form used in English for this unit, whether the cars come from Germany or Japan or Italy or wherever. Gene Nygaard

Note that my revert of "odd horsepower" was because the units were mixed in one article - kW first in some places and hp first in others. Also, it looked like you took a converted value for hp from kW and converted it back to kW, leading to precision errors. I'll take the discussion of which conversion to Horsepower, but think any discussion of mixing hp first and kW first belongs here instead. --SFoskett 14:04, May 31, 2005 (UTC)

Ah. I didn't realise that your term 'odd' meant inconsistent. Nor did I realise that you thought I had made conversion errors. I can explain both of those points. The values that I inserted came from manufacturer websites, not conversions. Sometimes the source had a hp value that differed from that already there.

For non-US vehicles the kilowatt value in official sources is often quoted as primary. The kilowatt values do not (as far as I know) vary by market. So that is convenient. The numerical hp values do vary by market. The hp values sometimes even vary within the market depending on whether it is a manufacturers press pack at launch or a manufacturers sales page at a later date. Where I could not find an official source, I made no change. The official information has eliminated some errors and improved consistency within varients. It is clear that in many cases, an engine has the same power in both markets as shown by the identical kW value. It is only the two forms of 'hp' that have different numerical values.

I hope that has given you some context. I think the official figures are an improvement. I don't like the inconsistency across models where I failed to find official values but I don't know what to do about that. Bobblewik (talk) 14:36, 31 May 2005 (UTC)

I don't know about this article, but it is the original measurement which should come first; there are likely articles in which that varies with different measurements. That is the consistency we should strive for, not a foolish consistency in listing the units of a particular system of units first. Often we have a source which lists numbers in two different units, with one of them carried to a ridiculous precision. That one is the converted value; the other the original. Sometimes it isn't that easy to tell; it often helps to guess at whatever you know about general practices at the time, and in the locateion, where the original measurement would have been made. Note that for engine power, if you have a several different numbers in one source given in two different units, and one of those units always has whole numbers divisible by five, or by ten, that is another indication to help determine which is the original. Gene Nygaard 20:20, 31 May 2005 (UTC)

I went ahead and edited my number mungler with this in mind. Now, when you enter PS (metric horsepower), it converts (correctly?) to hp and kW and optionally links those units to their correct wikipages. Hopefully if we start doing things like this 1001 PS (987 hp/736 kW) it will help solve this screwup. --SFoskett 21:06, May 31, 2005 (UTC)

Thanks for your comments. Although I have yet to use your number mungler, I am aware of it and appreciate your improvements in case I do use it. So far I have worked with official or at least respectable sources and the Google calculator.

As far as terminology is concerned, 'PS' and 'hp SAE' seem fine by me. I note that Morven said that hp sometimes is SAE gross and sometimes SAE net. So I am watching the debate with interest.

As far as original measures are concerned, I agree with Gene that the conversions should be secondary. The edits I have made are consistent with putting the raw value first. For example, I assumed that the best source for Mercedes Benz data was the German website. I could not find all engines specified but many were. They format power as 'kW [PS]' so I updated the article with that sequence and those numerical values. I happened to choose the term 'hp DIN' because I had seen it used elsewhere but the term could be changed easily enough. Bobblewik (talk) 12:32, 1 Jun 2005 (UTC)

I believe that the switch from gross to net hp happened after 1970 in the USA. So generally all pre-1971 hp ratings are gross and all post-1970 ratings are net. I have some car mags from the time period, and they note gross vs. net, but this practice ceased relatively quickly once all models switched over. --SFoskett 16:27, Jun 1, 2005 (UTC)

Note that "gross horsepower" is not terminology presently used in the article itself, and we have no idea what was done differently in measuring "gross horsepower" vs. "net horsepower". Gene Nygaard 16:35, 1 Jun 2005 (UTC)

[edit] Ambiguity in PS

I see that the article 'PS' says: Pferdestärke -- German for horsepower.

Is that correct? Bobblewik (talk) 14:23, 9 Jun 2005 (UTC)

Yes, that's what PS stands for. Generally, PS now equates to metric horsepower, while "hp" and "bhp" are more ambiguous. --SFoskett 15:46, Jun 9, 2005 (UTC)

Quite. I was wondering whether the simple statement needs qualifying in that article. Bobblewik (talk) 15:53, 9 Jun 2005 (UTC)

I don't think we need a whole lot on this disambiguation page, when it is discussed more fully in the linked article. However, it could be made a little bit clearer. I'm going to add "when PS is used as a symbol in English text, it's use is generally limited to the metric horsepower." Tweak it if you like. And maybe make the same point here in the horsepower article. Gene Nygaard 17:25, 9 Jun 2005 (UTC)

[edit] The Most Powerful Diesel Engine in the World

For information alone: The Most Powerful Diesel Engine in the World! --Lironos / 2LIRONOS7 02:04, 4 September 2005 (UTC)

[edit] POV

This article is very much centered around the US perspective. The "metric" horsepower is mentioned merely as an afterthought. Remember that the latter is in fact more broadly used worldwide. dsandlund 09:24, October 15 2005 (CET)

The English horsepower was first in time, by far.

The metric horsepower is a deprecated unit, not a part of the International System of Units and not acceptable for use with SI. And many are indeed going to primary measurements in the SI units, watts and kilowatts and megawatts. And the metric horsepower, when it is used, is most often not used with the "horsepower" name, even in English. In English, when it is used, it is usually used with the PS (or less commonly, CV) symbols. Gene Nygaard 09:13, 15 October 2005 (UTC)

[edit] hp (ECE)

"ECE is seen as slightly more liberal than DIN, and ECE figures tend to be slightly higher than DIN". I'd like to know where you found this little tidbit of information. I can quote you some contrary examples, that hp as measured by the ECE standard is actually lower than as measured by the DIN standard. In the late 80s and early 90s, French and Italian models "lost" about 2 or 3 hp on the conversion. The PSA Group and Renault still like to use hp (DIN) next to kW (ECE) as a marketing tool, such as the Renault Clio 2.0 RS (which was called Clio Sport 182 in the UK), a number which quoted the supposed 182 PS the car possessed, when in fact its power, according to the ECE standard, was 131 kW (178 PS). Pc13 16:45, 24 March 2006 (UTC)

[edit] Removed Link

This is my first time editing this talk thing. I hope I'm doing this right.

At some point, my horsepower article was added to the Wikipedia by someone who liked it, I guess. I know this from the referrer logs. I check the Horsepower article here occasionally, because I'm curious to see if my article is up to Wikipedia's standards. Recently, the link was removed by IJB TA, who stated "Removed link, inaccurate information, illustrations". That's fine, but I would be interested in knowing what is wrong, so it can be changed. It's not in my interest to be biased or appear smart, so if the whole thing is wrong, I will rewrite it completely without hesitation. While we are all here, however, I would like to know what is wrong so that the appropriate changes can be made. The link is here.

The illustration showing the power output of both engines shows almost identical power output above 400 hp, yet in the illustration of the powerbands for the engines the V8 somehow manages to have a much wider powerband. In the illustrations showing wheel torque the V8 lacks any dips for gear changes. A few things in the "A few things" section are not entirely accurate: A V8 would require a turbocharger of similar or even larger size to make the same amount of hp as a 4 cylinder [3]. It is the relationship of displacement to the size of the turbo that would make lag more prominent for the smaller engine. Turbo lag isn't usually a problem on a gear change during rapid acceleration. In your examples the 4 cylinder has a similar powerband and would not require more frequent shifts. There seems to be a few biased statements throughout the article, for example: "High revving engines will tend to be less reliable than low revving engines." that statement is highly circumstantial. The article has a slightly biased voice, but other than that it's very good. IJB TA 05:48, 16 October 2006 (UTC)

I have made substantial revisions to the article based on your input, and have re-added it to the wiki. Craigblock 22:43, 14 December 2006 (UTC)

Looks pretty good. Nice work. IJB TA 08:49, 11 January 2007 (UTC)

[edit] Metric Units

I see that the constant using kW, Nm and RPM as units of power has been changed from 7124 to 9553. I believe the correct value is 9549, as in (60*1000)/(2pi). I will change it to 9549 if people can verify that it is the correct value. —The preceding unsigned comment was added by Craigblock (talk • contribs) 17:43, 28 January 2007 (UTC).

[edit] Removal of text

I removed the following recently added text:

In the motor sports community it is a widely held misconception that the torque rating of a motor is an indication of how rapidly it will be able to accelerate a vehicle. Acceleration is work, and as such requires the application of force over distance (the definition of power). Horsepower or watts provide the indication of the power-producing capability of a motor.

First, acceleration is not work and force over distance is not the definition of power (it is the definition of work) and second, while horsepower is important for top speed, it is torque that determines acceleration. Consider the case of an electric powered vehicle at rest. At the moment that the vehicle begins to accelerate, the horsepower generated by the motor (no matter how powerful it is) is exactly zero. Alfred Centauri 02:44, 2 March 2007 (UTC)

You are very incorrect, a running engine (or electric motor) will never have zero horsepower. Because of how hp is measured (hp = tq x rpm / 5252) any rotating object can produce power. If you spin a bowling ball like a top and try to stop it with your hands it will put force on your hands at a certain rate, that is work being performed and to perform work you must have power. So even at idle you can imagine how an engine is making power. The thought that force starts the vehicle off and power keeps it moving is wrong because there is power at the very instant that motor begins to move. So in your electric car example the moment the car would begin to accelerate the motor would have a measurable amount of power. Also the reason power is so important to the performance of a vehicle (and the main thing many people don't understand) is that if you have more power you can create more force at a given rpm. To make a quick example, if you have an engine that makes 400 hp @ 10,500 rpm with 200 lb-ft of torque and an engine that makes 400 hp @ 5,250 rpm with 400 lb-ft of torque they will both be able to provide the same amount of wheel torque at the same rpm thanks to gearing.

200 lb-ft @ 10,500 rpm geared 4:1 = 800 lb-ft at 2625 rpm

400 lb-ft @ 5,250 rpm geared 2:1 = 800 lb-ft at 2625 rpm

The text you removed is not a misconception at all, the statement "horsepower is important for top speed, it is torque that determines acceleration." is very incorrect and a very common misconception. IJB TA 10:16, 11 March 2007 (UTC)

Sorry, but your words demonstrate that you are very confused. You are using words like power *very loosely*. Let me give you an example. You said "because there is power at the very instant that motor begins to move". This is clearly wrong. At the instant the motor begins to turn, it is by definition not turning so the power is exactly zero. This is simply by the definition. But hey, don't take my word for it. **Do the math**. To make it simple, assume the motor produces a constant torque, say 100 lb-ft, from the moment the acceleration starts until the motor reaches 1000 rpm, 10 seconds later. Since the torque is constant, the angular acceleration is constant (assuming losses are minimal). Thus, the rpm is increasing at the rate of 100 rpm per second. So, here is the result:

   Torque = 100 lb-ft from t = 0 to 10 (seconds)
   HP = 100 lb-ft * 100 * t / 5252 = 1.9 * t hp
   t = 0, hp = 0
   t = 1, hp = 1.9
   t = 2, hp = 3.8
   t = 3, hp = 5.7
   ...
   t =10, hp = 19

Don't you see? The horsepower is increasing every second yet the *acceleration* is constant!!! The torque is constant and the acceleration is constant. The horsepower is steadily increasing but the acceleration is constant. Hmmmm... Seems pretty obvious to me which one is responsible for the acceleration. Of course, if losses are *not* minimal but are increasing with rpm, then the acceleration will not be constant but will be reduced as the speed ramps up. However, the result is the same.

Once again, (and this is basic physics - don't take my word for it, go look it up!) it takes (net) torque to change angular speed. This is the definition of torque. Forget any of the other things you have heard - without net torque, angular speed doesn't change. And, if angular speed doesn't change, there is no linear acceleration developed by the tires on the road. To change the angular speed from zero doesn't require any power, only torque. To maintain speed, requires power only if there are losses. For an automobile traveling down the highway at high speed, there are losses from air resistance, tire to road friction as well as mechanical losses in the drivetrain. Thus, to *maintain* this high speed requires power (and therefore torque) from the engine. But, to *increase* that speed rapidly requires far more torque than horsepower.

Consider the following problem that I would think is more applicable to your mind set. Let's say that a car weighing 3000lbs is moving at 60 mph and engine is turning 2000rpm. Let's say the wheels are rotating at half that, 1000rpm (this implies that the radius of the wheel/tire combo is about 10 inches (the diameter would be about 20 inches). The losses (air resistance, tire friction, internal friction, etc.) require that the engine produce 10hp in order to maintain that speed. Using your hp formula above, the torque produced by the engine is 26.26 lb-ft (10hp *5252/2000).

Now, let's say we want to accelerate this vehicle, at the rate of 10mph/per second. This means we need to increase the angular speed of the engine by 334 rpm per second or the wheel/tire rpm by 168rpm per second. Well, at the instant the acceleration starts, the engine is still turning 2000rpm but our torque is way up. In order to *accelerate* the car, we need to change the wheel/tire rpm by 168rpm per second, which requires and additional torque of 1385 lb-ft at the wheels or 692 lb-ft from the engine. That's in *addition* to the torque required to make up the losses at 60mph. So, take a look

To accelerate from 60mph at the rate of 10mph/sec we need:

   Torque = 719 lb-ft which is constant
   Acceleration = 10mph/sec (actually a poor assumption since the air resistance goes up as the square of the speed!) starting at t = 0
   t = -1, hp = 10
   t = 0, hp = 273
   t = 1, hp = 319
   t = 2, hp = 364
   etc...

So, there ya go. To get that acceleration, we're going to need that torque. Note that, once again, the hp is going up as time progresses but the acceleration is remaining constant. Hmmmm... Alfred Centauri 02:15, 12 March 2007 (UTC)

Yes I understand that it is wheel torque that determines the acceleration of a vehicle. The point is that in order to make more wheel torque you must make more power at the flywheel. That is why the statement "horsepower is important for top speed, it is torque that determines acceleration." is so very incorrect. A vehicle with more power will be able to accelerate faster than a vehicle with less power because it can provide a greater amount of force at the wheels. Power to weight ratio, hmmm...wonder why that came about instead of torque to weight ratio. IJB TA 02:33, 12 March 2007 (UTC)

No, to make more wheel torque requires more torque from the engine. You still don't 'get it' You are using the word power when you should actually be using torque. Please consult a physics text to learn the difference. Or, try this for a second opinion: http://g-speed.com/pbh/torque-and-hp.html Alfred Centauri 02:43, 12 March 2007 (UTC)

You're forgetting the something VERY important, GEARS! Gears multiply torque. Again:

200 lb-ft @ 10,500 rpm geared 4:1 = 800 lb-ft at 2625 rpm

400 lb-ft @ 5,250 rpm geared 2:1 = 800 lb-ft at 2625 rpm

If you do the math you will notice both of those examples have 400 hp. IJB TA 02:53, 12 March 2007 (UTC)

IJB, when you're right you are not addressing the points Alfred is making. And, when you do address them, you're wrong. Your grip on what the terms torque, power & work mean in physics is poor. Paul Beardsell 03:27, 12 March 2007 (UTC)

So my math is wrong? Please feel free to point out my mistake. IJB TA 03:31, 12 March 2007 (UTC)

I didn't say your maths is wrong. Paul Beardsell 03:35, 12 March 2007 (UTC)

Well that's the basis for everything I'm saying here, so how can it be wrong? IJB TA 03:41, 12 March 2007 (UTC)

You say that torque is this and power is that. You are told that your understanding is fundamentally incorrect. And this is explained, and reference material is suggested. You ignore this but reply by saying (I paraphrase) "Look, 2x6=12 and 3x4=12, so therefore I am right." I say there's nothing wrong with your maths, but your physics could do with a re-grounding in the basics. Paul Beardsell 03:52, 12 March 2007 (UTC)

I was trying to be as basic as possible. If you want more torque at a given rpm you need more power, if you want more rpm at a given amount of torque you need more power. How much more basic can you get? IJB TA 03:57, 12 March 2007 (UTC)

Whoa! That's just plain confused! If, as you said, you want more rpm at a given amount of torque, you just plain screwed. Without more torque, you will never, ever get more rpm. Plain and simple. If you disagree with this most basic result, you're just being dense.

Second, you said "If you want more torque at a given rpm you need more power". Well, that's a funny way of saying something that is just plain obvious. After all, the equation says power = torque * rpm. Thus, more torque = more power or more power = more torque for a constant rpm. It's not that you need anything at all, it's just what happens. That's what the equation says and, as far as I know, no one disagrees with that equation. So, what is the point that you are actually trying to make here? 72.148.204.187 04:39, 12 March 2007 (UTC)

Here, let's try this out. If I have a car with 10,000 lb-ft of torque and a car with 100 lb-ft of torque which one accelerate faster? IJB TA 03:46, 12 March 2007 (UTC)

This must be a trick question but I'll bite. Since you said nothing of torque peaks etc, I'll assume that the engines produce their max torque at the same rpm. Further, I'll assume the cars are identical in every way except for the torque production of the engine. Now, assuming that the *wheel* speed is the same for both cars (both cars are traveling at the same speed), the car with 10,000 lb-ft of torque will vastly out accelerate the car with 100 lb-ft of torque.

Of course, you could cook up something like this: for the first car, the wheels turn 10 revolutions for each revolution of the engine and for the second car, the wheels turn 1 revolution for each 10 revolutions of the engine. In that case, the torque delivered to the wheels is the same for both cars so, assuming the first car is traveling 100 times slower than the second car, the acceleration would be the same for each car (ignoring all other factors such as air resistance, tire resistance, etc). 72.148.204.187 04:48, 12 March 2007 (UTC)

I'll answer the question but only if you demonstrate that you know the physics. Paul Beardsell 04:08, 12 March 2007 (UTC)

You said That is why the statement "horsepower is important for top speed, it is torque that determines acceleration." is so very incorrect. Unless and until you have enough physics to understand why you are wrong there is no point in entering into a discussion. There are a hundreds of basic physics textbooks, some of them online, that you could consult. Torque is force x (perpendicular distance from the centre of rotation). You should know that F=ma (Newton 2) so a is proportional to F. Or "it is torque that determines acceleration". Power is the amount of work done per second. Work is the distance moved against the net force. Speed is distance per unit time. So "power is important for top speed." Paul Beardsell 04:08, 12 March 2007 (UTC)

I know the physics well enough, I've already demonstrated that. IJB TA 04:18, 12 March 2007 (UTC)

So you agree with the paragraph above (starting "You said" and ending "top speed")? Paul Beardsell 04:24, 12 March 2007 (UTC)

This is part of what has been cut from the article: Acceleration is work, and as such requires the application of force over distance (the definition of power). No. Acceleration is not work. Acceleration is the rate of increase of speed. Or rate of increase in the rate of increase of distance. the second derivative of distance i.r.t. time. Power is the rate at which work is expended. Work, not power, is "the application of force over distance". Incorrect text cannot appear in an encyclopedia article. Paul Beardsell 04:22, 12 March 2007 (UTC)

No, I don't agree and you didn't answer my question. IJB TA 04:31, 12 March 2007 (UTC)

Look, this is so easy to see, I can't believe that there is even an argument over this. OK, take the case of an engine that produces its peak torque of 1000 lb-ft at 3000 rpm. Above 3000 rpm, the torque linearly declines until is only 500 lb-ft at the engine redline of 6000 rpm. With me so far? What is the power output at torque max? HP @ Torque_max = 1000 * 3000/5252 = 571 hp. What is the max hp and at what rpm does it occur? That's easy... max hp is 643 @ 4500 rpm.

So, to get maximum acceleration, you want to be geared such that, at your *current* wheel rpm (could be small, could be large), the engine is at 3000 rpm. So, if your current wheel speed is small, you can put much greater torque to the wheels than the engine can produce because of the torque multiplication from the transmission. If you wheel speed is fast, well, you can't get as much torque to the wheels because, your gear ratio is smaller. Still, under max acceleration at this point, the power output can be no larger that 571hp and thus, the power delivered to the wheels cannot be greater than (and is less than due to losses) 571hp. That's because, no matter how hard you try, a ***transmission cannot create power*** - power in >= power out.

But, what about maximum speed? For that we need to know the forces on the car at high speed but let's say that they go like this.

Speed Power Loss

60mph 40hp

120mph 160hp

180mph 360hp

240mph 640hp

OK, so to go 240mph requires a constant power production from the engine of at least 640hp. To get this much horsepower, we need the engine to be turning 4500rpm. If we have 18" wheels, they will be turning 4500rpm so our gearing has to be 1:1 at this point. Obviously, if our gear ratio is greater, we'll never hit that speed but that's due to the red line, not a lack of power. Point is, the maximum speed is determined by the max hp, not the max torque. At max torque, we're only going to produce 571hp and, **no matter what kind of gearing you use**, your top speed will be limited to 226 mph. Period, end of story. 72.148.204.187 04:25, 12 March 2007 (UTC)

Yes a transmission only transfer power, hence the name. Part of the point here is that flywheel torque is made meaningless by the transmission. IJB TA 04:31, 12 March 2007 (UTC)

Really? So, if flywheel torque is meaningless then it becomes a free variable. I can set it to zero and not affect the torque to the wheels. Riiiiggghhhttt.... Are you sure you want to make such a statement as that? Further, what does you 'point' mean in the context of my example? Specifically, how does your 'point' change my example results? Do you agree with my results? If so, don't you agree that the max hp set the upper limit on speed, not the max torque? If you don't agree, please point out the error. I don't see any connection between my example and flywheel torque or how that might pertain to the qeustion of the correctness of the results. 72.148.204.187 04:58, 12 March 2007 (UTC)

Sorry, I should have said it's meaningless by itself, you still need an rpm rating to indicate how much work can be performed. Once you have a torque rating with an rpm rating you might as well just call it power so you can use one number instead of two that say the exact same thing. Ever wondered why Watt came up with horsepower in the first place? IJB TA 05:14, 12 March 2007 (UTC)

[edit] Removal of text (reprise)

Alfred removed the following text because it is wrong.

Incorrect material cannot be allowed in the encyclopedia.

Paul Beardsell 04:36, 12 March 2007 (UTC)

Please explain how this is incorrect. IJB TA 04:39, 12 March 2007 (UTC)

First sentence: I have no idea what the "motor sports community" believes but if they believe as alleged then they are correct.

Second sentence. Already addressed. I repeat:

This is part of what has been cut from the article: "Acceleration is work, and as such requires the application of force over distance (the definition of power)". No. Acceleration is not work. Acceleration is the rate of increase of speed. Or rate of increase in the rate of increase of distance. the second derivative of distance i.r.t. time. Power is the rate at which work is expended. Work, not power, is "the application of force over distance". Incorrect text cannot appear in an encyclopedia article. Paul Beardsell 04:22, 12 March 2007 (UTC)

Third sentence. Not incorrect but it doesn't tell us anything. Of course the unit of power indicates the power. It's just like saying "Inches or centimetres provide the indication of the distance."

Paul Beardsell 04:55, 12 March 2007 (UTC)

Corrected the text. IJB TA 04:51, 12 March 2007 (UTC)

Hooray! I'll have a look and report back. Paul Beardsell 04:55, 12 March 2007 (UTC)

No better. Reverted. Paul Beardsell 04:58, 12 March 2007 (UTC)

Fine I'll run a new version by you here.

Among automotive enthusiasts - Better?

I don't care how you phrase it. But when we get the physics sorted out I will ask you to provide citations. Paul Beardsell 05:17, 12 March 2007 (UTC)

it is a widely held misconception that the torque rating of a motor is an indication of how rapidly it will be able to accelerate a vehicle. - Definitely a common misconception.

No! You are wrong. Your motor sport buddies are correct. Paul Beardsell 05:17, 12 March 2007 (UTC)

The torque rating of a motor (flywheel torque) does not, isn't that what I just showed here? IJB TA 05:43, 12 March 2007 (UTC)

Acceleration requires work, - What is the difference between work and power?

I thought you said you understood the physics? I have already said what the difference is. As has Alfred. It's all written here, above. Paul Beardsell 05:17, 12 March 2007 (UTC)

and as such requires the application of force over time (the definition of power). - Define power please.

I already have done so. As has Alfred. He and I have pointed out that "force over distance" is the defintion of work, not power. It so happens that "force over time" (also known as "impulse") is equivalent to work. Neither is the defintion of power. Paul Beardsell 05:17, 12 March 2007 (UTC)

Horsepower or watts provide the indication of the power-producing capability of a motor. - Looks right to me. IJB TA 05:07, 12 March 2007 (UTC)

It's right but loosely worded enough to make me think that you believe it only provides an "indication". Not only an "indication" - the number of hp or W is the power. Not just an indication. Paul Beardsell 05:17, 12 March 2007 (UTC)

Incorrect text will not be allowed in the article. Paul Beardsell 05:17, 12 March 2007 (UTC)

Horsepower can only provide an indication of performance because of the large number of variables involved in comparing vehicles, torque alone on the other hand can't provide an indication with any accuracy. IJB TA 05:25, 12 March 2007 (UTC)

No! Torque tells us about acceleration. Just like your motor sports enthusiasts say. Paul Beardsell 05:36, 12 March 2007 (UTC)

You're right, this is pointless: Horsepower or watts provide the indication of the power-producing capability of a motor.

Horsepower or watts provide the indication of the performance capability of a vehicle. - Better? IJB TA 05:25, 12 March 2007 (UTC)

When you say "the indication" it can be read as "the only indication". I would allow "an indication". And, of course, there are many "performance capabilities": Braking, cornering, endurance, ground clearance. But for any material to be in the article it must make sense contextually. It isn't enough for the material to be correct. It must fit properly into the structure of the article. The material must be encyclopedic. It must tell us something true and relevant and it must cite its sources. Paul Beardsell 05:31, 12 March 2007 (UTC)

Horsepower or watts provide an indication of the acceleration performance capability of a vehicle. - Better?

I could elaborate it more later. My main concern is that it is correct so that you wont remove it again. Will you remove it? IJB TA 05:40, 12 March 2007 (UTC)

What power provides is speed. Torque provides acceleration. This, I was starting to hope, you would now accept. If you add incorrect material I will remove it. Paul Beardsell 05:42, 12 March 2007 (UTC)

How many different ways do I have to say/show this? How do you make more wheel torque to make a vehicle accelerate faster? POWER! How do you make a vehicle with a higher top speed? POWER! Why is that so hard to understand? IJB TA 05:48, 12 March 2007 (UTC)

Anyway, I'm sorry if you don't understand it, but that's not my problem. IJB TA 05:56, 12 March 2007 (UTC)

It wasn't my intention to wear you down. I am sorry I haven't succeeded in persuading you why you are wrong. Consider this: The max torque and max power of a engine is typically at different RPM. At which RPM do you get best acceleration? At the max torque RPM or at the max power RPM? Paul Beardsell 06:21, 12 March 2007 (UTC)

At peak wheel torque. Now why there is a difference in your mind between making this wheel torque with a large amount of torque at low rpm or a small amount of torque at high rpm at the flywheel is what I don't understand. IJB TA 06:32, 12 March 2007 (UTC)

Oh, you are being deliberately disingenuous! At which ENGINE RPM does PEAK WHEEL TORQUE occur? Max power (engine) RPM or max torque (engine) RPM? Ask any racing driver, at which ENGINE RPM does max acceleration occur? At max power (engine) RPM or max torque (engine) RPM? Paul Beardsell 06:41, 12 March 2007 (UTC)

Max wheel torque will be at max flywheel torque. That should be obvious! IJB TA 06:48, 12 March 2007 (UTC)

So far so good. And so why is max torque (flywheel) RPM different from max power (flywheel) RPM? Paul Beardsell 06:56, 12 March 2007 (UTC)

It doesn't matter. An engine could produce peak power and peak torque at the same rpm. IJB TA 06:58, 12 March 2007 (UTC)

But, typically, they do not. So, answer the question. Paul Beardsell 07:02, 12 March 2007 (UTC)

Irrelevant. IJB TA 07:07, 12 March 2007 (UTC)

So now Howstuffworks.com is wrong too? So the person that wrote that text originally, Howstuffworks.com and myself are all wrong? Where is a reference that supports what you're saying? IJB TA 06:37, 12 March 2007 (UTC)

Y'know, it is possible that Howstuffworks.com is wrong. If it is wrong I bet I'll get the author of the material to admit it quicker than you. Gimme the exact reference and I'll go have a look. I'm betting you have misinterpreted the material. Paul Beardsell 06:41, 12 March 2007 (UTC)

How likely is it that Howstuffworks.com is wrong? A car is considered to be "high performance" if it has a lot of power relative to the weight of the car. This makes sense -- the more weight you have, the more power it takes to accelerate it. For a given amount of power you want to minimize the weight in order to maximize the acceleration. Hmmm...don't see anything about torque in there. IJB TA 06:45, 12 March 2007 (UTC)

Written for kids. No one here has not said there is a relationship between power and torque. The relationship is one you should know. But acceleration is directly proportional to torque, not to power. Your motoring buddies are correct: The acceleration of a vehicle is best judged looking at the torque figures, not the power. Who do you trust better? Isaac Newton or some kiddie article at Howstuffworks.com? Paul Beardsell 06:50, 12 March 2007 (UTC)

I trust the math. Yes it is written so anyone can understand it, but that doesn't mean it's wrong. It talks about the relationship of torque and hp at the beginning. IJB TA 06:54, 12 March 2007 (UTC)

No it doesn't. Following your link the article I get to does not mention torque. Paul Beardsell 07:04, 12 March 2007 (UTC)

Go to the first page. IJB TA 07:12, 12 March 2007 (UTC)

Who said Newton was wrong? He was very right, you're just now understanding what power actually is. Motoring buddies? IJB TA 06:56, 12 March 2007 (UTC)

F=ma. Yeah, y'know, the ones who you claim are wrong about torque. So, c'mon: Max acceleration at max torque RPM or at max power RPM? Paul Beardsell 07:02, 12 March 2007 (UTC)

AGAIN! Transmission! What does it do to torque? You can multiply, can't you? IJB TA 07:06, 12 March 2007 (UTC)

I'll provide the reasonable answer you refuse to give. Every racing driver knows max acceleration occurs at max torque engine RPM. Torque provides acceleration. F=ma. Newton 2. Sure, engines with larger torque output are more powerful than other engines at the same RPM. And engines which rev faster are more powerful than others with the same torque. But acceleration depends on torque. Paul Beardsell 07:13, 12 March 2007 (UTC)

Yes,

200 lb-ft @ 10,500 rpm geared 4:1 = 800 lb-ft at 2625 rpm

400 lb-ft @ 5,250 rpm geared 2:1 = 800 lb-ft at 2625 rpm

Now which engine would accelerate the same vehicle faster? Which engine makes more torque at the flywheel? Which one makes more torque at the wheels? IJB TA 07:17, 12 March 2007 (UTC)

Just because you ignore it does not mean it isn't true. IJB TA 07:43, 12 March 2007 (UTC)

Are you complaining about the half hour delay while I had supper? Paul Beardsell 08:10, 12 March 2007 (UTC)

Now we're in the position of Archimedes prepared to move the earth given the right lever and fulcrum. Because the 0 to 10 km/h acceleration of a bulldozer is better than that of a F1 racing car. Why? Good traction and low gearing. How come you go on and on about power being more important than torque when you've got to consider the weight of the vehicle (F=ma) and the composition of it's tires (or tracks!) to work out how quickly it can really accelerate? A massively heavy vehicle with a 5hp lawnmower engine will, with the right gearing, out accelerate any racing car 0 to 0.001 km/h. It's all down to torque. Maximum speed, on the other hand, is down to power. Paul Beardsell 08:10, 12 March 2007 (UTC)

Now we have the para jointly crafted by ourselves in the article, and it seems correct, let's ask this question: Is it relevant? Perhaps we should just cull it. Paul Beardsell 08:10, 12 March 2007 (UTC)

nowhere to find PS JIS, a substitude to HP in car engines.

I still question whether it's accurate to say torque at all. It's not torque all by its lonesome self that's performing the work to accelerate something. Force still has to be applied at a certain rate to have acceleration. In your bulldozer example the bulldozer might be able to accelerate faster because of gearing, not torque, it still takes power to apply force at a certain rate. Electric motors will produce their peak torque at 0 rpm, but if you try to accelerate an electric car with the brakes on there will be a whole bunch of torque there but nothing will happen, you still need motion. Once you have motion and force you have power. In your lawn mower example you might be able to make it happen like you said but if you look at what's really going on you will notice that it would only happen because of the way the F1 car is geared, and it would not be true just because of torque alone. For example: a 5 hp lawnmower engine geared 1,000,000:1 can provide ~7,290,000 ft-lbs of torque at 0.0036 rpm, a 750 hp F1 engine geared 35,217.4:1 can provide ~7,290,000 ft-lbs of torque at 0.5395 rpm. So if you were to drop the clutch of both of these crazy powertrains you can see how the one with more power would accelerate faster because it's able to apply the same amount of force at a higher rate. You could have literally all the torque in the world and not be able to accelerate something with it if it's not applied at a certain rate, when you have a rate of force you have power. Power performs work, not torque. IJB TA 21:12, 12 March 2007 (UTC)

[edit] Added disputed tag

Rather than start an edit war, I have placed a disputed tag for the following text:

The power output of a motor is an indication of how rapidly it will be able to accelerate a vehicle. Acceleration requires force or torque. The acceleration of a vehicle is directly proportional to the torque delivered to the wheels. (See Newton II.) The amount of torque delivered to the wheels is dependent on the power output of the engine as well as the efficiency and gearing of the powertrain. The maximum speed of a vehicle also depends upon the engine's power output. [3][4]

The first sentence and fourth sentences are incorrect. This is not an opinion but is a fact that follows directly from the math. The vehicle acceleration is given by:

$acceleration = \frac{wheel\ torque}{wheel\ radius * vehicle\ mass}$

The wheel torque is the engine torque multipled by the gear ratio. Note that power output does not come into this equation at all. The power output of the engine is not an indicator of the acceleration. Alfred Centauri 19:56, 12 March 2007 (UTC)

One last thing, what happens if you put the formula for power in the equation above? You get:

$acceleration = \frac{wheel\ power}{wheel\ rpm * wheel\ radius * vehicle\ mass}$

Look! Now there is another variable in there that wasn't before - rpm. What happens when you set the wheel rpm to zero? The equation blows up (acceleration goes infinity) unless something else is simultaneously true - the wheel power is also zero. But if, as the disputed sentences claim, that acceleration depends on power, how would a car ever get started when there is clearly zero power at the wheels? If the disputed statements are true, a car would never accelerate from a standing start! Alfred Centauri 20:19, 12 March 2007 (UTC)

You're going all chicken and egg here. If you don't have power you don't move, if you don't have force you don't have power. The point here is, if you have more power you can accelerate faster because you can apply more force at a greater rate. IJB TA 21:28, 12 March 2007 (UTC)

Utter nonsense. We've gone over the zero speed start before and you're just being dense if you don't get it now. However, regarding your last statement, you need to come to grips with the fact that the force is the torque. Don't you understand this? Torque is the linear force multiplied by the radial distance from the axis of rotation. There is no power involved here. Torque is (force * lever arm). Do you see a power term in there anywhere??? Applying force at a greater rate is equivalent to saying applying torque at a greater rate if the lever arm is unchanged. You reaalllyyy don't have a good grasp of these words do you? What is key here, it turns out (see my new info below), is to achieve matched action. The maximum acceleration is obtained when the action of the engine equals the action of the wheels. However, this can only occur at one speed in a fixed gear. As the speed increases (or decreases) away from this point, the acceleration is driven more by the engine torque than by its power. If you have a bunch of gears, then you can match the actions at several different speeds. But, that's a whole different story. Unless this information is provided in the article, the statement that power determines acceleration is simply wrong. Alfred Centauri 21:58, 12 March 2007 (UTC)

You need to come to grips that power is the rate of force. You need a measurement of time (rpm) to have motion. You do know that acceleration happens over time don't you? IJB TA 22:36, 12 March 2007 (UTC)

Wow! Power is the rate of force??? I'll throw away all my text books now, the're clearly all wrong when they say power is the rate of change of energy. I guess the're also wrong about rpm being a frequency which is not a measurement of time. And no, you don't need an increment in time for motion. Have you ever taken a calculus class. Does the phrase 'instantaneous velocity' have any meaning to you? Alfred Centauri 22:55, 12 March 2007 (UTC)

Sorry, I'm in a hurry. So maybe this is more accurate: Acceleration occurs over time, as far as I know instant acceleration would require infinite force. I don't believe vehicles built by humans are capable of that. Acceleration requires work, to perform work you must have power. Am I correct? IJB TA 23:00, 12 March 2007 (UTC)

If by instant acceleration, you mean an instantaneous change in speed, then you are correct, that would require an impulse (an infinite force applied for 0 time). However, I haven't mentioned that anywhere. Further, you are correct in that to change an objects velocity, there must be work involved. So, let's see how good your intuition is. Let's say I accelerate one object from 0mph to 10mph over 1 second using a fixed force. At the same time, I accelerate another object that was moving at 10mph to 20mph with the same fixed force. Same force, applied over the same length of time to identical objects with the only difference being their initial speeds. Is the amount of work done the same for both objects? If not, which one required more work What is the power required to start the first object moving? What is the power required to start the second object accelerating? Alfred Centauri 00:02, 13 March 2007 (UTC)

[edit] Torque, power & acceleration: Two misunderstandings

Misunderstanding #1 (Alfred's?): Power = RPM * torque. Acceleration is directly proportional to (wheel) torque. But power is directly proportional to both RPM and torque. The power of an engine is therefore an indication of the acceleration potential of the vehicle. Two identical vehicles with engines with the same max torque, but generated at different engine RPM, will generate different torque at the driven wheels at the same wheel RPM. Different RPM with same torque means different power. Paul Beardsell 21:19, 12 March 2007 (UTC)

I think I understand that equation very well. Bottom line, the power required for a particular acceleration depends on the current speed. The torque required does not. See my new info below. Alfred Centauri 21:37, 12 March 2007 (UTC)

Misunderstanding #2 (IJB's?): Power = RPM * torque. But torque itself depends on RPM. Internal combustion engines generate their maximum torque at a given RPM. Electric direct current engines generate their max torque at zero or very low RPM. Maximum power is always at higher RPM than the RPM of max torque. Max vehicle acceleration always occurs at the max torque engine RPM, not max power RPM. Paul Beardsell 21:19, 12 March 2007 (UTC)

Where in the rpm range an engine produces its peak torque is not even relevant here. One vehicle will still accelerate faster if it has more power over its entire rpm range compared to another one. If you do the math you will notice that is true because the vehicle that produces more power over its rpm range will also be able to provide more force over the whole rpm range compared to the other engine. IJB TA 21:28, 12 March 2007 (UTC)

[edit] New info on hp versus torque debate

The following is a quote from this web site: http://www.allpar.com/eek/hp-vs-torque.html

OK, so what about power? As has been noted by a previous contributor, Power (hp) = Torque (ft-lb) * RPM / 5252. Note that power is also force * velocity, specifically: Power (hp) = Force (lb) * Velocity (MPH) / 374 That's net horsepower, which is engine power minus losses like transmission and tire friction. The force is the sum of the longitudinal forces at the contact patches of the two driven tires. Hmmm... P = F * V ...rearrange to get F = P / V ... that means that you get the maximum force pushing the car if you maximize your *Power* at any given velocity. This gives us another useful rule: * Shift to maximize engine POWER, not engine torque!

This is interesting, because I think this might point to the heart of our debate as to which determines acceleration. While I have taken into account the effect of the transmission, I have mostly thought about this problem in terms of a fixed gearing. In other words, Assuming you are in *a* gear, what determines instantaneous acceleration and the answer is without a doubt engine torque. For the same *fixed* gear ratio for two different engines, the engine producing the most torque produces the most acceleration.

But what about the opposite side? What about an ideal continously variable transmission that allows the engine to remain at a fixed rpm. Here's the question: which rpm do you set for the engine to get to some speed as quickly as possible? It turns out, that for the continously variable transmission, the answer is that you set the rpm for maximum power rather than maximum torque. You have to solve a differential equation to get this but here's the result:

$velocity = 2 \sqrt{\pi \frac{engine\ power}{vehicle\ mass}\ * t}$

This assumes a standing start at t = 0.

So, in this case, the acceleration is determined by the max power output of the engine. The reason this works out this way is the the continously variable transmission acts analogously to a continously variable turns ratio for a transformer always matching the load (the ratio of the wheel torque to wheel rpm) to the source resistance (the ratio of the engine torque to the engine rpm).

This matching ensures that the power delivered to the wheels is the maximum possible and is constant with time. With a fixed gear, this matching occurs only at one speed. At any other speed in this gear, the condition of max engine torque gives the best acceleration.

If you have five gears, you can match at 5 different speeds, 100 gears, 100 different speeds, infinite gears match at all speeds which is why the continously variable transmission gives best acceleration with max power rather than max torque.

Bottom line is this, the statement that power determines acceleration is strictly true only when the following is true:

$engine\ torque\ *\ wheel\ rpm\ =\ wheel\ torque\ *\ engine\ rpm\$

When the products above are vastly different, torque determines acceleration.

Shall I edit the article accordingly? Alfred Centauri 21:34, 12 March 2007 (UTC)

You're still not getting the point. It is not always true that a car with 500 lb-ft of torque will accelerate faster than a car with 200 fl-lbs of torque. Saying more engine torque makes a car accelerate faster is lacking information, it's like saying the Sears tower is bigger than both of the Petronas towers. There's just not enough there to make the statement true. Engine/flywheel torque alone does not make a vehicle accelerate faster. IJB TA 21:50, 12 March 2007 (UTC)

You're correct, it's not always true. In fact, that is precisely what I just said in my comments above. Didn't you read them? Oh well, here's a brief summary. Torque primarily determines acceleration when the engine action and wheel action are grossly unmatched (such is the case with the zero speed start). Power primarily determines acceleration when the actions are matched or almost matched.

However, unless there is a continously adjustable transmission keeping the actions matched, the actions will match at only speed per gear. Alfred Centauri 22:07, 12 March 2007 (UTC)

That's kind of true. But there's maybe a fraction of a second where the engine with more low rpm power has the advantage. Granted that could make or break a drag race but it still doesn't explain the majority of time the vehicle is accelerating and it's still lacking information. What if the engine with more torque only produces it a 10 rpm and another with 200 lb-ft at 10,000 rpm. If you say power on the other hand all the information you need is right there, the RATE that torque can be applied. IJB TA 22:32, 12 March 2007 (UTC)

Really? Take a look at this image: [4]. This is the torque and power curve for a typical DC motor. Looking at the power curve, I would say that there ain't much power near zero shaft rpm. Yet, that is precisely where the most powerful acceleration occurs given a constant gearing. Alfred Centauri

Oh, and once again, power is NOT the rate at which torque changes. They have the same units but, they are not the same thing. By elementary calculus, the rate of change in torque is F * r' + F' * r. In words, it is the tangential force times the rate of change of the lever arm + the rate of the change of tangential force time the lever arm. Further, consider the case where a changing torque is applied to an immovable flywheel. The torque 's a-changin but there's no power at all. No work being done, nada. Power is NOT the rate of change of torque. Alfred Centauri 01:05, 13 March 2007 (UTC)

Look, I don't want to sit here and throw a whole physics book around with you and I don't have time to. The torque available at the flywheel of an engine does not indicate the performance of the vehicle, PERIOD. I don't see how anything you are saying here proves that torque alone makes a vehicle accelerate faster. If it were as simple as that a torque to weight ratio would be set in stone. Go calculate the torque to weight ratio for a bunch of sports cars and see what happens. Be sure to present you results here. IJB TA 01:34, 13 March 2007 (UTC)

How much work can you do with 500 ft-lbs? Answer that one without some kind of measurement of time. IJB TA 01:43, 13 March 2007 (UTC)

IJB said: "Look, I don't want to sit here and throw a whole physics book around with you and I don't have time to." I don't either and besides, it isn't necessary. We certainly don't need electromagnetics, optics, special relativity and quantum mechanics to resolve this debate. The physics behind this don't take a book but only a page. So, your statement doesn't make any sense excepting the case that you really don't know how to intepret or debate the results I have provided so you are looking for an 'out'. Further, you keep asking odd things like "go calculate the torque to weight ratio". Why? Then you ask what amount of work I can do with 500 ft-lbs. Are you aware that 500 ft-lbs IS an amount of work? Work is measured in ft-lbs. On the other hand, so is torque BUT, despite the fact that the units are the same, they do not have the same meaning. Are you aware that work is a scalar quantity but torque is a vector? Power is a scalar quantity too but rate of change of torque is a vector. They are not the same thing.

I come up with a valid point and then you counter it with some odd statement or question that tells me you don't have any idea what I just said. Too bad...

By the way, the answer to your question is (assuming you mean 500 lb-ft of torque) this: The amount of work you can do with 500 lb-ft of torque is an element of the positive real numbers. I don't know how else to answer that and further, I don't quite see how your question or its answer is relevant to this debate. If you point is simply that torque alone does not tell us anything about the work that can be done, I agree. Likewise, power alone tells us nothing about the work that be done. Once again, what IS your point? Alfred Centauri 02:11, 13 March 2007 (UTC)

My point: FLYWHEEL TORQUE DOES NOT INDICATE THE PERFORMANCE OF A VEHICLE. IJB TA 02:15, 13 March 2007 (UTC)

[edit] Unnecessary Debate

There are external links right in the horsepower article that show how acceleration is affected by an engine's power or torque figures in clear terms. I don't see any debate here. The debate tag is gone and everyone can go home. 74.101.94.34 01:56, 13 March 2007 (UTC)

The statement "The torque delivered to the wheels (which are rotating at a certain angular velocity) is directly proportional to the engine's power output" is clearly untrue. What if the certain angular velocity is zero or near zero? What if the maximum torque of the engine is developed when the shaft speed is zero (as is the case for a DC motor)? According to equations in the article, power is proportional to angular speed. If the angular speed is zero, power is zero. Consider a DC electric motor that is stalled because some nitwit left the parking brake on in his golf cart. What is the power produced by the engine? Zero. What is the power delivered to the wheels? Zero. Yet, the motor is clearly producing torque at the shaft and at the wheels. How do you reconcile this situation with the statement that "The torque delivered to the wheels is directly proportional to the engine's power output? Clearly, this statement, in its unqualified form is incorrect. Alfred Centauri 02:21, 13 March 2007 (UTC)

The case where the angular velocity is zero does not prove anything. When the angular velocity is zero, there is no work, no power, no time, and obviously not _enough_ torque to be doing anything. It is similar to standing on a diving board. There is torque but no motion. No matter how anyone spins this, the hard fact that the acceleration of a moving object will always be a=P/mv, not a=F/m. Why? Because the latter is for statics and does not consider dynamics. The reason that torque can be used to _determine_ acceleration on a car is because statics can be used only at that instant. Gears can be used to increase torque, at the expense of revs, relative to the _power_. 74.101.94.34 03:02, 13 March 2007 (UTC)

What??? Where do you come up with such nonsense??? Gee whiz guys (or gals), this isn't rocket science! How can you possibly say such nonsense as there is no time when the angular velocity is zero? Are you saying that the world comes to a stop when we stop at a red light??? (I know that isn't what you meant but what the heck else does 'no time' mean?)

Further, a = P/mv is an indeterminate form when v = 0. Are you saying that, in a dynamic situation, v never goes through zero??? Have you ever seen a harmonic oscillator? What is a when v = 0 there? Certainly not a static situation, yet a=F/m gives the answer right away. Nope, can't use a = P/mv in general. Besides, P is a derived quantity which is a function of velocity so a=P/mv is dividing a function of velocity by the velocity which is why you get the weird indeterminate form. F is not a function of velocity (in general) and, ignoring special relativity, neither is m. So, a = F/m is more fundamental and more general. Not valid for dynamics??? Boy that's a good one. Newton's rolling over. Maybe your thinking of something else like F = dP/dt where P is the momentum (mv), not power. I dunno what's going on here. It's like stepping into the twilight zone. I don't think I've heard such hogwash outside of usenet. You folks ever heard of calculus? Alfred Centauri 03:45, 13 March 2007 (UTC)

Oh, and one other thing. a = P/mv only when E = 0.5mv^2. Look at the energy of a harmonic oscillator - it's 0.5mv^2 + 0.5kx^2. So, the power P = mva + kxv which gives a = (P/mv - kx/m). Hmmm... don't look like your equation at all! However, a = F/m still works since F is just -kx (F don't depend on no stinkin velocity!). Alfred Centauri 04:05, 13 March 2007 (UTC)

I really don't appreciate your tone. If you want to believe that torque adds kinetic energy, then go ahead. Fact is, in mechanics, an object will gain kinetic energy (as in: change velocity, or accelerate) at the rate the work is being done. The rate of work is power, PERIOD. You obviously have faith in your "views" on this part of physics and therefore cannot be touched. Please just don't screw up the Wikipedia article on horsepower with your nonsense. I'm done with this. Hopefully innocent readers will not be affected by the garbage that can be found on this talk page. 74.101.94.34 04:14, 13 March 2007 (UTC)

I apologize for my tone - I got carried away with this debate. However, my tone does not in any way diminish the truth with what I have said here. I'm not making this stuff up. It's just elementary physics. What you have said above is true. The rate of work is power and a change in kinetic energy requires work. Nonetheless, a = P/mv is just not generally true. Does this seem contradictory to you?

Look once again at the harmonic oscillator. The acceleration of the mass is, at all times, -kx/m. The term P/mv is exactly zero for all time. Why? The power associated with the mass is zero at all times. Why? As the mass gains kinetic energy, it looses potential energy so that the total energy of the mass is constant with time. Thus, a = P/mv fails for this case. Perhaps you think that I'm spewing garbage here so please, take the time to independently verify this information. You can start right here on Wikipedia. But before you do, would you please do me the favor of:

pointing out where in any of my comments where I have asserted that torque without angular displacement adds kinetic energy.

pointing out where in any of my comments I have asserted that work is not required to accelerate an object.

pointing out where in any of my comments I have asserted something that is not physically true according to any undergraduate physics text.

The fact is, unless I have made a typo somewhere, there is nothing that I have asserted that contradicts classical physics.

Finally, the only edits I have made to this article have been to remove questionable material and to put a disputed tag on the material when it was reinserted. Do you really want to say that is 'screwing up' and article? Alfred Centauri

[edit] A suggestion in the hopes of removing the dispute tag

First, I hope we can agree with this equation for the acceleration of the car:

$a=\frac{\tau_e\ r_g}{R_w\ m}$

This says that the acceration of the car is equal to the engine torque times the gear ratio divided by the wheel radius times the mass of the car.

Clearly, the acceleration is directly proportional to the torque developed by the engine at that instant. The more torque the engine can produce at that instant, the more acceleration there will be at that instant.

Second, I hope we can agree with this equation for the rate of change of kinetic energy of the car:

$\frac{dKE}{dt}=\frac{d}{dt}(\frac{1}{2}mv^2) = mva$

Now, use the acceleration from the first equation in the second equation to get:

$\frac{dKE}{dt}=\frac{\tau_e\ r_g}{R_w}v = P_e$

This says that the rate of change of KE at any instant is equal to the engine power output at that instant. That's no surprise, its just conservation of energy at work.

OK, the key thing to get out of this is the following: The instantaneous acceleration is not NOT proportional to the instantaneous power output of the engine but is instead proportional to the instantaneous torque developed by the engine. I've used the word instantaneous here three times for emphasis because I beginning to think that this whole debate may be due to a loose use of the word acceleration.

When I use the word acceleration, I mean the rate of change of velocity and, as I've showed above, that is not proportional to power. However, if by acceleration, someone is thinking of something like the shortest 1/4 mile time, then I can see why someone would say that is determined by power. Obviously, to reduce the 1/4 mile time requires more average power. I completely agree with that statement.

So, here's the statement I propose:

The vehicle acceleration at any instant (how far you are pressed back in your seat) is proportional the engine torque at that instant. The average vehicle acceleration over some period of time is proportional in part to the average power output of the engine during that time. However, to maximize average power, it is not enough to choose an engine with the highest power output since the maximum power output occurs at one particular rpm. So, it is also important to choose the gear ratios wisely and to shift the gears in such a way that, on average, the engine is close to its maximum power output most of the time.

I'm going to go ahead and edit the article with this statement. Alfred Centauri 16:10, 13 March 2007 (UTC)

No. The vehicle acceleration at any instant ... is proportional to the ~~engine~~wheel torque at that instant. Paul Beardsell 02:11, 15 March 2007 (UTC)

Please don't. It isn't appropriate on an article about horsepower. Horsepower is simply a unit of power. Dissertations as to the best way to maximise acceleration of a vehicle should happen elsewhere at WP. Rather at power, or at acceleration, or at racing car but not at horsepower. Similarly, this is not appropriate material for the article Watt, or some other article about a unit of power. Paul Beardsell 20:57, 13 March 2007 (UTC)

I agree with Paul here. It just causes too many problems because of the way people interpret the text. Obviously as Alfred put it, The instantaneous acceleration is not NOT proportional to the instantaneous power output of the engine but is instead proportional to the instantaneous torque. But it's not proportional to the torque developed by the engine because the amount of torque at the wheels will always be different due to gearing and the less than perfect efficiency of the powertrian. So obviously instantaneous acceleration will be proportional to the amount of torque at the wheels not the engine. My only point was that you can provide more torque to the wheels at a higher rpm if more power is produced by the engine, that's all. I wasn't trying to debate the laws of physics here. I was just hoping a widespread misconception could be cleared up here, but as Paul said this isn't really the place to be doing any kind of rumor control. So I vote for complete removal. IJB TA 21:59, 13 March 2007 (UTC)

Then by all means, please remove the text entirely. In fact, that is exactly what I did in the first place. This entire debate took place because the material was reinserted.

By the way IBJ, do you understand what proportional means? It appears from you statement above, that you do not. You said "it's not proportional to the torque developed by the engine because the amount of torque at the wheels will always be different due to gearing...". Nonsense. Two things are proportional if they are related by a multiplicative constant. Gearing only changes the constant.

And, therefore, changing the gearing destroys the proportionality! So acceleration is NOT proportional to engine torque in a geared vehicle. That is IJB's point. Paul Beardsell 02:14, 15 March 2007 (UTC)

Yes I agree that losses destroy the simple proportionality but, as long as the delivered torque is large, the losses are insignificant and simple proportionality is a good approximation. Heck, if you going to be picky, what about the losses between the torque applied to the wheel by the axle and the linear force applied by the tire to the road?

On second thought, I'll just remove the text now. Alfred Centauri 23:33, 13 March 2007 (UTC)

"Yes I agree that losses destroy the simple proportionality but..." ummm...isn't that what I said? Some vehicles have much more efficient powertrains than others. I think you're just purposely ignoring my point only because I don't word everything exactly the way you would. That's extremely annoying and completely unnecessary. (WARNING! I WILL BE USING WORDING THAT I KNOW TO BE ACCURATE, I APOLOGIZE IF IT'S NOT TO YOUR TASTE) Unless of course you're actually saying that one vehicle with more power will not be able to provide more torque to the wheels at a higher rpm than one with less power if all other things are the same except gearing. Will an engine absolutely need to produce more torque at the flywheel to make more torque at the wheels at the same (or higher) rpm compared to an engine with less torque? IJB TA 00:36, 14 March 2007 (UTC)

It's not about taste and I'm sorry to have annoyed you but when you make a blanket statement about something, don't you think you ought to know the meanings of the words you use? It is my opinion that you often do not. For example, you just said: Some vehicles have much more efficient powertrains than others. Don't you know that efficiency is a proportionality constant? Percent loss does not destroy proportionality. Gearing and efficiency are mutliplicative constants that do not destroy proportionality. Didn't you say that in your previous comment? Let me see, yes here it is:

But it's not proportional to the torque developed by the engine because the amount of torque at the wheels will always be different due to gearing and the less than perfect efficiency of the powertrian. So obviously instantaneous acceleration will be proportional to the amount of torque at the wheels not the engine.

Yep, that's what you said. Acceleration is not proportional to engine torque because of gearing and less than perfect efficiency. There it is, plain as day. You said that and it's just plain wrong.

Screw efficiency - we all agree losses are negliglible for the purposes of the argument. It is not proportional BECAUSE OF GEARING. Paul Beardsell 02:17, 15 March 2007 (UTC)

Here's a test... Which of the following wheel torques are not proportional to the engine torque? Which are approximately proportional to wheel torque? Which are proportional to wheel torque?

a) wheel torque = gear ratio * engine torque

b) wheel torque = gear ratio * %efficiency * engine torque

c) wheel torque = gear ratio * %efficiency * engine torque - 100 lb-ft

d) wheel torque = gear ratio * %efficiency * engine torque - 10 lb-ft

??? Alfred Centauri 01:52, 14 March 2007 (UTC)

They're all different gear ratios, my dear Alfred. Paul Beardsell 02:17, 15 March 2007 (UTC)

See there you go again. Why do even bother posting a response? I'm wrong about something other than my main point, whatever, I don't care. Now are you going to actually answer my question or are you just going to waste both of our time and nit-pick everything I say so you can avoid giving a response that would indicate that the power output of the engine indicates the acceleration performance of the vehicle? IJB TA 02:28, 14 March 2007 (UTC)

I'm sorry, I thought we were talking about your main point. Oh I see, my bad. You said: "My only point was that you can provide more torque to the wheels at a higher rpm if more power is produced by the engine, that's all". Well, I do agree with that.

I must say that this debate has prompted me to look into this much deeper than I ever have and for that I thank you. I was preparing to make some charts showing torque, power, and acceleration versus engine speed, then wheel speed, then time at several different gear ratios. These charts would, I believe, clearly show that the maximum instantaneous acceleration of the vehicle always occurs at torque maximum yet the maximum average acceleration occurs in the vicinity of the power maximum. Then, I would let the number of gears get arbitrarily large and show that, not surprisingly, fastest times to speed occur if the engine is kept at the max power rpm constantly while the gear ratio changes continously. I may still do that but I'm afraid I've got to prepare the test for my EE students. 72.148.204.187 02:53, 14 March 2007 (UTC)

No, max acceleration happens at power maximum because max wheel torque happens at max engine power rpm not at max engine torque rpm. Paul Beardsell 02:17, 15 March 2007 (UTC)

This discussion has definitely made me realize that I should get back in school to improve my understanding of math and physics. I just wanted to clear up a commonly used and commonly misunderstood saying, I didn't expect it to be so involved. I definitely learned a few things from your posts, I thank you for that. IJB TA 03:59, 14 March 2007 (UTC)

Sorry if I'm jumping into the conversation late, much of I what I'm about to say might be a repeat of what others have said, you have been warned ;-) Power, torque, and RPM are all related, I think we can all agree on that. Engines produce power, I think we can all agree on that. Neglecting friction, given any amount of power, I can produce an arbitrarily high torque using a transmission, I think we can all agree on that. Now, let's say I've got a lawn mower engine producing 3.5 HP. I can gear it down and produce more torque than the fastest production car made today. I think we can all agree that a 3.5 HP lawn mower engine is not going to yield good acceleration compared with a 200 HP engine. You need power, power is no more than the rate of energy. Energy is conserved, torque is not.

Here's another way to look at it: what do you need to climb a hill at high speed? Hint: a big engine with a lot of *power*. A small engine geared way down will produce as much torque as you like, but at the expense of RPM (Power is more or less conserved). The end result is lower power means low acceleration no matter how you cut it. There's no such thing as a free lunch.

Now for the final straw. A car at velocity v has well known kinetic energy:

$E = \frac{1}{2} m v^2$

If, at time t1, the velocity is v1 and at time t2, the velocity is v2, the difference in energy is:

$\Delta E = E_1 - E_2 = \frac{1}{2} m v_1^2 - \frac{1}{2} m v_2^2$

The definition of power is the time rate of change of energy, so the average power is:

$P_{avg} = \frac{\Delta E}{\Delta t}$

$P_{avg} = \frac{m (v_1^2 - v_2^2)}{2 (t_1 - t_2)}$

Notice that RPM, torque, force, etc. do not appear in any of the above equations. You can derive the equations using those terms, but they will cancel out leaving only mass, velocity, and time. Let's say I have a 1000 kg car (about 2200 lbs) and I want to accelerate from 0 to 60 mph (about 27 meters/sec) in 6 seconds. That will require a minimum of about 61 kW or about 81 HP. Those numbers are close to the performance of a 1986 Mustang GT. Of course that Mustang had a 302 cu. in. (5.0L) engine that could produce a maximum of 200 HP at about 4800 RPM or so. On average it produces much less than 200 HP because it doesn't operate a peak power during the 0-60 acceleration. A hybrid electric drive (as used by diesel-electric locomotives, ships, or submarines), could squeeze a lot more power out of the engine and radically increase acceleration performance. Of course, you probably new that. Madhu 04:21, 14 March 2007 (UTC)

(sorry to butt-in here but I wanted to make sure it was clear who I was replying to) Nice summary, Madhu. Bottom line, the energy required to change speeds depends not only on the change in speed, but also on the average of the final and initial speeds:

$\Delta E = E_1 - E_2 = \frac{1}{2} m v_1^2 - \frac{1}{2} m v_2^2 = m(v_1 - v_2)\frac{v_1 + v_2}{2}$

That is, it takes much more energy to change from 50mph to 60mph than it does to change from 0mph to 10mph.

This has always struck me as odd. If a apply a constant force to an object, the acceleration is constant, yet the power (flow of energy into the object) increases linearly with the velocity: P = Fv. Odd. Just goes to show that using power to determine acceleration is tricky. To determine acceleration via power alone requires knowledge of the current velocity which is the time integral of the acceleration. Thus, to determine the acceleration you must 'know' the past history of the acceleration! On the other hand, the acceleration at any instant is proportional to the net force at the same instant. Much easier, don't you think? Alfred Centauri 19:35, 14 March 2007 (UTC)

::Oh, I just noticed something you said that needs to be qualified. You said "The end result is lower power means low acceleration no matter how you cut it.". That's not true unless you are using acceleration to mean something other than dv/dt. As I've just shown, talking about power and acceleration without any mention of the current speed is tricky. Think of it this way, I can make the acceleration arbitrarily high with an arbitrarily low amount of power. Your final equation says so.

$\hat P = \frac{m (v_1^2 - v_2^2)}{2 (t_1 - t_2)} = \frac{m \hat v \hat a}{2} ==> 1W = m \hat v 50g ==> \hat v = \frac{1W}{50gm}$

Do you see? Reducing the power doesn't limit the acceleration but instead, limits the average of the final and initial speeds: (v1 + v2)/2. Alfred Centauri 21:13, 14 March 2007 (UTC)

Sorry, my bad. You did qualify it a couple of sentences earlier ("at a high speed"). I agree, at a given speed, lower power <=> lower acceleration. Alfred Centauri 22:04, 14 March 2007 (UTC)

I think we are all in violent agreement with each other ;-) I remember reading that a human on a bicycle can out accelerate a car at very low speeds, that's sounds consistent with what you are saying. I'm coming at it from the angle of minimizing 0-60 time or 1/4 mile time. The above analysis (if I did it right) is based purely on conservation of energy, so I believe a minimum of 81 HP is required to accelerate a 1000 kg car from 0-60 in 6 seconds. If it can be done with substantially less average power, I'd like to know how. I reserve the right apply the mean value theorem ;-) Madhu 02:14, 15 March 2007 (UTC)

Torque is just how an engine makes power. Strap a rocket to the back of a car. Is there any torque? 74.101.94.34 16:02, 14 March 2007 (UTC)

The answer is yes. In this case, the linear force of the rocket is applied (via the road surface) to the tire. The torque on the wheel is simply this force times the radius of the wheel/tire combo. If the wheels are free to spin, this torque will cause the wheels to spin up. If the brakes are on, the brakes will apply an opposing torque to keep the wheels from spinning.

The answer expected in any reasonable physics exam, where the mass of the wheel would be assumed zero, is that there is no torque on the wheel unless the wheel brakes are applied. So I correct the above para: "The answer is no." Paul Beardsell 02:22, 15 March 2007 (UTC)

By the way, I think I have come up with a nice summary of what we can say about the instantaneous acceleration of a vehicle as it relates to the various power and torque measures.

Here's the setup: imagine two vehicles with identical mass, wheel size, and powertrains but with different engines. Neglecting external loss factors, the following statement is always true:

At any instant in time, the vehicle producing the greatest wheel torque has the greatest acceleration (always true!).

It is also true that:

At any instant in time, if the vehicles have the same gearing, the vehicle producing the most engine torque has the greatest acceleration.

It is also true that:

At any instant in time, if the vehicles have the same speed, the vehicle producing the most wheel power (or equivalently engine power) has the greatest acceleration.

It is also true that:

At any instant in time, if the vehicles have the same speed and the same gearing, the vehicle producing the most engine torque (or equivalently, the most engine power) has the greatest acceleration.

So, when the speeds and gearing match, the engine producing the most torque is also producing the most power so choose either one to get the best acceleration.

When the speeds match but the gearing does not, the engine producing the most power wins out.

When the speeds don't match, we can only go by wheel torque unless the the gearing is matched in which case we can go by engine torque too.

Alfred Centauri 18:36, 14 March 2007 (UTC)

I think this site says it best: "The fastest accelerating vehicle is the one that has the highest average force going to the pavement through a wide range of speeds." IJB TA 23:29, 14 March 2007 (UTC)

It sounds OK to me except for the phrase "fastest accelerating". I'll bet that you know exactly what that phrase means, but frankly I don't. That's because I'm accustomed to using certain words precisely as it's a necessity in my profession. In my world, acceleration is large or small, positive or negative, constant or changing, but never fast or slow. So, here's what I think; if 'fastest acceleration' means 'shortest time to change speed' then I think it's right on the money. Alfred Centauri 00:34, 15 March 2007 (UTC)

(Note: I wrote this in reply to Paul's earlier post which he has modified (now apparently deleted: What's going on Paul?) resulting in an edit conflict, oh well). Since wheel torque = engine torque * gear ratio and since you have given me an infinitely variable gearbox, I can make the torque infinite at any engine rpm where the engine torque is non-zero (and maybe even then if you consider that infinity * zero is an indefinite form). Perhaps you meant a continuously variable gearbox? If so, then we want to have the engine producing maximum power at all times. That gives us the greatest power to the wheels so we can change our energy at the maximum rate available to us (which of course is when we have the engine developing maximum power). Does this give us the largest torque at the wheels. No. Does this gives us the largest average torque at the wheels? Yes. Why? It can be no other way. The fastest time to change speed must give the largest average acceleration: (v1 - v2)/(t1 -t2). Since acceleration is proportional to the torque at the wheels, the largest average acceleration can only be due to the largest average torque at the wheels. Do you agree? Alfred Centauri 01:23, 15 March 2007 (UTC)

Largely nonsense. You cannot make the torque infinite given a particular non-zero wheel speed. Of course I mean continuously, you are being unnecessarily pedantic. Yes it bloody well does give the largest torque at the wheels (for the current speed). You are wrong! Not just the average, always the largest. All this precision is hiding understanding. The largest possible average acceleration happens when instantaneous acceleration is at its max over the range. Always. And you have already allowed me an infinitely variable gearbox (with no wheel slip). Paul Beardsell 01:59, 15 March 2007 (UTC)

Paul, we're in violent agreement. The largest torque at the wheels (for a current wheel speed) occurs when the engine is producing maximum power. Further, I agree that the largest average acceleration occurs when the instantaneous acceleration is maximum over the range. Alfred Centauri 02:04, 16 March 2007 (UTC)

I think we need agree on what we are trying to achieve. I am talking about low 0-60 MPH or 1/4 mile times. Instantaneous acceleration is something else, and usually not interesting in a pedestrian sense. Madhu 02:14, 15 March 2007 (UTC)

At the drag strip, max acceleration happens at max engine power RPM, not at max engine torque RPM. Paul Beardsell 02:34, 15 March 2007 (UTC)

Paul, please take a breath and try to tolerate these questions from me. By 'max accleration', I assume you mean 'max average acceleration'. If not, and if the dragster doesn't have a continuously variable gearbox, then I disagree with that statement. Here's a question for you. I would be grateful if you would take the time to read the question and answer it to the best of understanding:

A vehicle with a manual 5-speed transmission can be equipped with one of two engines. Both engines produce the same max horsepower at the same rpm (286hp @ 5000 rpm). However, one engine produces a peak torque of 600 lb-ft @ 2500 rpm and the the other engine produces a peak torque of 310 lb-ft @ 2500 rpm. Which of the two engines would give the max average acceleration in the 1/4 mile? Alfred Centauri 02:04, 16 March 2007 (UTC)

[edit] The penny drops, acceleration, power and torque

If an engine generates its max power, say 100 units at 5 rpm, because power=torque x rpm, engine torque at 5 rpm must be 20. If the engine produces its maximum torque, say 40 units of torque at 2 rpm, then power at 2 rpm is 80. Because we are going to use a continuously and infinitely variable gearbox (doesn't everybody?) (or we are going to choose the optimal gearing for each example), the max wheel torque happens at max engine power RPM (apparently something commonly disputed in the crowd at motor races). Why? Because max power rpm gearbox ratio for a given speed will be 2/5ths of the ratio necessary at the same speed for max engine torque rpm. If max wheel torque at max engine RPM is 20, then at max engine torque RPM the wheel torque will be 40*2/5=16. Max acceleration happens at max engine power RPM, not at max engine torque RPM.

That isn't just a happy choice of numbers. Max power happens at a higher RPM than max torque (as it always does, otherwise max power would be at max torque RPM) then

(torque at max power)*(max power RPM) > (max torque)*(max torque RPM)

Were this not the case, we get:

(power at max power RPM) < (power at some other RPM)

which is a contradiction.

Sorry it took me so long to get this. As a saving grace I point out that this is the simplest explanation - it's all understandable without integration, calculus, or a discussion of work or energy. Just Newton 2. Paul Beardsell 01:38, 15 March 2007 (UTC)

To make it plain: Max wheel torque is at max engine power RPM, not at max engine torque RPM. Paul Beardsell 01:53, 15 March 2007 (UTC)

(Warning: there was another edit conflict so I don't know if my response is restating the obvious or what!)

Huh? Max wheel torque occurs at max engine power rpm? Nonsense! With a continuously variable transmission (and assuming we actually want to get to speed in the shortest time), max wheel torque occurs at our minimum speed. It has to be that way, Paul. The power delivered to the wheel is (ideally) constant. Let's avoid the case where the wheel speed is zero for the moment and simpy say our continuously variable transmission must, at all times, match the wheel speed to the constant (max power) engine rpm. Thus, the torque gets larger as our wheel speed gets smaller.

Look, I think I know what you are trying to say here but be careful. The fully qualified statement is: Given a continuously variable transmission and a given wheel speed, the maximum wheel torque is developed when the engine is developing maximum power. But, forget putting numbers in for a second. Look carefully at the setup:

At all times, our transmission matches our wheel speed to some constant engine speed so our wheel torque goes up as speed goes down.

If our engine speed to set to max power, we can say that the power at any particular wheel speed is greater than that for any other engine speed setting.

That's all we really need to know. This condition ensures that the acceleration at any particular wheel speed is greater than for any other engine speed setting. But, this is counterintuitive! It seems that the maximum acceleration should occur when the engine is producing maximum torque! But look:

If, at the same wheel speed, our engine rpm is set to max torque, the gear ratio is clearly less than in the case when our engine rpm is set to max power (max torque rpm < max power rpm). Thus, although the engine is producing maximum torque, the torque at the wheel is actually less because the gear ratio is smaller than when the engine is set to max power rpm.

Yes, I realize that this is what you said, but I thought I would rephrase it to see if we actually do agree on this! Alfred Centauri 02:44, 15 March 2007 (UTC)

IJB's understanding was right all along, just his language was sloppy. You add unnecessary complication/obfuscation. Of course, if I have a continuously variable gearbox I am going to use the thing to provide optimum acceleration. I did say the same speed. You start off by saying "Huh? Max wheel torque occurs at max engine power rpm? Nonsense!" Now I don't care if in the remaining paras you resile from this or not and I am not going to even read it. It's not a matter of pride, just of time. I repeat: Max wheel torque occurs at max engine power RPM. Plain and simple. Knowing that, vary your geartrain ratio accordingly. Paul Beardsell 02:59, 15 March 2007 (UTC)

Yes, I do see that you described the gear ratios later in terms of the same wheel speed. So, as I said, I know what you were trying to say. But, yell it as loud and as often as you like and it won't change the simple fact that your statement is wrong in general without qualification and its easy to demonstrate this.

Max wheel torque for a fixed gearing occurs at max engine torque rpm.

Max wheel torque for a fixed wheel speed occurs at max engine ~~torque~~ power (ooops!) rpm.

So, do you still believe that your statement is true in general? If you choose to take this personally and not read the rest that's your choice and not in my span of control. So be it. Where you see obfuscation, I see clarification. Alfred Centauri 04:00, 15 March 2007 (UTC)

It's difficult not to feel aggrieved when "later" is the self same paragraph or, more to the point, someone spends half a dozen paragraphs agreeing with everything you have said but starts off with "Nonsense. You are 100% wrong." And then, having comprehensively agreed with you, next post continues to state you are wrong. I state it again, loud: Max wheel torque occurs at max engine power RPM, not at max engine torque RPM. Of course the same wheel speed is assumed, that's the understanding when such statements are made. We don't need to say same vehicle, same altitude, same fuel composition, same wheel diameter. We say what varies, not what remains the same. Different RPM and same wheel speed necessarily implies a different gear ratio. What is necessarily implied does not need to be stated. You would have us say "Given that the basic laws of physics do not change over the duration of the experiment...". I too make that assumption, I just don't feel it necessary to state it. Paul Beardsell 04:45, 15 March 2007 (UTC)

I agree that IJB's understanding was right. And, if he had used more precise language, such as the quote he provided, there would have been no debate from me. Do you think that in an encylopedia such statements as his original one have a place? Is an encyclopedia an appropriate place for sloppy language and unqualified generalizations? Do you not believe that we all gain insight by carefully examining what we say and strive to make it more precise? Alfred Centauri 04:00, 15 March 2007 (UTC)

No, here we agree. I argued with him on two counts. Firstly his use of physical terms incorrectly (I was right) and on his understanding (I was wrong). As, I believe, you did also. Sobering, isn't it, when two people who consider themselves usually correct on issues such as this, two who can walk the walk and talk the talk, are so wrong on the basic understanding, especially if one of them is a lecturer in a closely related subject at a university, and still has it wrong. Paul Beardsell 04:45, 15 March 2007 (UTC)

Not so fast, Paul. I'm ready to admit to an error in reasoning, but, to be honest, I still am not sure what it was he was trying to say. The problem is, he can't describe to me his understanding in words that I will interpret in the same way he does. Thus, he might be saying something that I entirely agree with but I won't know unless I question him thoroughly. If he says something that I believe is clearly erroneous, I will say so and I will explain why I think so in terms that he may not interpret the way I do.

But, rather than repeat once again what I think, please see my question to you about two engines. I think the answer will surprise you. In my opinion, the answer contradicts what, to the best of my understanding, you and LBJ are saying. That's why I would appreciate you answering the question using what you believe is the correct reasoning. Then, I'll answer it with my reasoning and then at least, we'll know exactly each other's position is on this. Alfred Centauri 02:24, 16 March 2007 (UTC)

Alfred said: "Max wheel torque for a fixed wheel speed occurs at max engine torque rpm." Wrong:

Engine characteristics: max power 100 @ 5 rpm (torque 20), max torque 40 @ 2 rpm (power 80).

Wheel speed 10rpm. Wheel torque @ engine 5 rpm = 10; Wheel torque @ engine 2rpm = 8.

Paul Beardsell 04:59, 15 March 2007 (UTC)

I completely agree with you, what I wrote is wrong. That's what I get for being careless with my copy and paste of the previous statement. I've corrected my previous post. Now, are we in agreement? Alfred Centauri 02:09, 16 March 2007 (UTC)

[edit] Two impossible engines

If we had two ideal engines, one producing constant torque and one producing constant power, it's easy to show that either one could outperform the other. Of course the constant torque engine is theoretically capable of producing arbitrarily high power and similar for constant power. The reality is that piston engines are neither (they tend to be closer to constant torque, but drop off at the low and high ends). The problem is even more acute for gas turbines, their torque is relatively low, so their RPM is relatively high (20,000 RPM or more) at peak power. Operating the engine at a constant speed (max power if you are not concerned with durability) is usually the most optimal solution for high performance. This technique is used by diesel-electric locomotives, large ships, submarines, larger propellor driven aircraft, generators, helicopters, and someday soon, automobiles. At the end of the day, engines are not well suited as prime movers over a wide range, but electric motors are. I wrote up a short article on this last year. Madhu 02:53, 15 March 2007 (UTC)

In my field of EE, an ideal constant torque engine would be analogous to an ideal constant voltage source (or a constant current source depending on your choice of flow variables). Ideal sources contain infinite energy. Typically, a source resistance would cause a drop in voltage due to load current but, a gasoline engine actually produces more torque (up to a point) with increasing load current so, at first blush, it would appear that it's source resistance is negative (up to a point) and then turns positive after the torque peak.

Yes the ideal source contains infinite energy, but that doesn't mean we can't model real world sources as near ideal sources for the purposes of analysis (EE is my field also). Regulated power supplies are quite close to ideal, they just limit the maximum power they can deliver. Piston engines exhibit a fairly flat torque for a significant RPM range. There is a simple justification for this: if the throttle is held constant, a fixed amount of fuel/air mixture enters the cylinders on each cycle (the displacement does not vary with RPM). That fixed charge of fuel/air produces a fixed pressure upon ignition. That fixed pressure produces a fixed force on the piston, as it has constant area. The force on the piston produces a fixed torque on the crankshaft as the crank arm length is constant. Of course, there are many variables, such as variation in fuel/air flow rate, but that's a secondary effect. Whatever negative slope exists in the torque curve apparantely is not significant. Madhu 18:32, 15 March 2007 (UTC)

However an ideal constant power engine is something I haven't thought about. Let's see, an electrical constant power source would produce a voltage proportional to the root of the load resistance or a current proportional to the root of the load conductance. Hmmmm.... I wonder if I could use a FET in the feedback path of an op-amp? Gotta go now but thanks for the stimulating comment! Alfred Centauri 04:12, 15 March 2007 (UTC)

This is why I bring up diesel electric drives. It's a constant speed engine, a regulated power supply electrically. Usually, these engines are operated at peak power RPM (or close to it). The simple reason is that it maximizes power/weight ratio. It's not exactly a constant power engine, as that would try to push power into a load that might not want it, but it can deliver constant power over a wide range of wheel speeds. Madhu 18:32, 15 March 2007 (UTC)

[edit] A choice of two engines

Engine 1 has a way wider power band. 600 lb-ft @ 2500 is 286hp, so presumably it's making way more average power than the engine 2. Engine 2 only has 148hp at 2500. 74.101.94.34 21:38, 16 March 2007 (UTC)

True, engine 1 produces more or less flat power (286 HP) between 2500-5000 RPM. As long as you keep engine RPM in the this wide range (not hard to do), you are getting to most power available. Engine 2 produces 286 HP in a much narrower range, so it's hard to get max power out of it. Of course, if we have a low loss continuously variable drive train, the results will be more or less equal. Madhu 00:06, 17 March 2007 (UTC)

Here's another one: engine 3 produces a peak torque of 63 lb-ft @ 35000 RPM and weighs the same as engine 1 and 2. You have your choice of drive train. Which engine produces the highest average acceleration in the 1/4 mile? Madhu 00:06, 17 March 2007 (UTC)

Engine 3 makes 419 horse. It is the best no matter what the application (reliability, cost, etc aside). All that matters is the power. 74.101.94.34 00:27, 17 March 2007 (UTC)

Please, Madhu! This is my problem ;<). Seriously though, I'd like make the point that, despite the max horsepower of these engines being equal at the same rpm, the average acceleration produced is vastly different. And, the difference between the engines is the max torque rating. Yes, engine 1 has (by design) constant power output between 2500 and 5000 rpm, while engine 2 has more or less constant torque over the same range.

The point I'd like to make is that a poor drive train can skew the results. Even with 1/10th the torque of engine 1, engine 3 is the better choice. If you haven't guessed, engine 3 is an Allison 250-B17C. Madhu 01:25, 17 March 2007 (UTC)

Now, a lot of debate has been going on about what is more important, torque or horsepower, when it comes to acceleration (which almost everyone involved has interpreted to mean a different thing!). What can we learn from this example? Alfred Centauri 00:35, 17 March 2007 (UTC)

IMHO, all you need is lot of power and a continuously variable drive train (all else being equal) ;-) If you have a 10th century drive train, your milage may vary. The whole debate of torque vs. power is like asking which is better 120 Volts @ 20 Amps or 240 Volts @ 10 Amps? Same thing, different transformer. Madhu 00:53, 17 March 2007 (UTC)

Yes, I think we all have come to the conclusion that, given a continuously variable transmission, power is king. But, assuming that isn't case, what is more important: torque, power, or ... something else. That is, if all you have is a choice of 2, 4, 8, 16, & 32 ohm taps on your transformer, and you want to get the most current through a purely inductive load in the least amount of time, what's more important on the source side: voltage, current, or... Alfred Centauri 02:01, 17 March 2007 (UTC)

The laws of physics do not change with respect to transmission availability. People tried to explain to you long ago that acceleration of a moving object is dependent on the power. 74.101.94.34 02:26, 17 March 2007 (UTC)

Do you really believe what you just said is true; that acceleration is dependent on power? a (acceleration) = F (force) / m (mass). Where do you see power anywhere in that equation, my friend? Is this equation only true for non-moving objects? Where does the max instantaneous acceleration occur within a fixed gear? (Hint: it's not generally when the engine is producing max power). Alfred Centauri 13:21, 17 March 2007 (UTC)

I am not sure what your point is. Yes, force (or torque) working against mass produces motion. Power can produce force (or torque) at some rate. They are all related, I can derive any one from any other. At zero speed (or RPM), power is not delivered to the load, just as a short circuit dissapates no power (at the short), but that doesn't mean power is not produced. If you want to know max instantaneous acceleration, I would suggest deriving the equations of motion (not easy to do with a non-uniform power curve) and differentiate as needed. Madhu 14:55, 17 March 2007 (UTC)

There are two points I'm trying to make to anonymous. If I'm not mistaken, anonymous has claimed in a previous post that a = F/m is only applicable to static situations and that a = P/mv is always applicable. So, the first point is that a = F / m is valid always. My second point is that the acceleration of an object is not dependent on power for three reasons: (1) v is not a constant in this equation so a depends on the ratio of P/v. In fact, a = P/mv is a differential equation! (2) The P here is the rate of change of kinetic energy, mva. Thus, P/v is simply F and we're back a = F/m. (3) P is not always just the rate of change of kinetic energy so, in general, a = P/mv doesn't hold. Alfred Centauri 16:00, 17 March 2007 (UTC)

OK, for the discrete transmission, based on your previous example, I'll say it's an engine with a constant power band that's wide enough to cover all of the discrete gear ratios. The continously variable transmission is just the limit case where the power band can be made arbitrarily small. Another way to look at: given the width of the power band, choose the number and value of gear ratios to maximize engine power. I think 74.101.94.34 stated that earlier. You'll have to be more specific in the case of the inductive load. Madhu 02:32, 17 March 2007 (UTC)

Bingo! It's not max power (alone) nor max torque (alone) that determines average acceleration in the case of a finite number of gears. It's the power bandwidth. Let's call it the 'Q' of the engine if you will. So, to the best of my understanding, the following are statements of fact:

For a finite set of fixed gears:

maximum instantaneous acceleration (dv/dt) within any gear occurs at the engine max torque rpm.

maximum instantaneous acceleration (dv/dt) for a given wheel speed occurs at engine max power rpm. (This will less than or equal to a_max at max torque within any gear).

the engine producing nearly constant power over a wide range of rpm will produce the maximum average acceleration (vf - vi)/(tf -ti)

Do you disagree with these statements? (hopefully I've avoided any typo's this time!) Alfred Centauri 13:21, 17 March 2007 (UTC)

It's a little confusing because I'm not sure which variables are constant and which ones are not. If the first statement is basically

F = m a

T = I ω

, I agree, I'm not about to argue with Netwon. The second one seems to be some kind of restatement of conservation of energy, which I won't argue with either. If you mean to say

F = m a

or energy is conserved, I'd rather just say that instead of making a highly qualified statement that depends on the readers interpretation, unspecified power curve, or gearing. Madhu 14:43, 17 March 2007 (UTC)

It sounds like you are looking at the reverse problem rather than the forward problem. The reason we have multiple gears is because we are talking about a powerplant with narrow powerband. If we had a powerplant with a wider powerband, we wouldn't need so many gears. With all due respect, your statements seem overly complicated to me, but what do I know ;-) Madhu 14:43, 17 March 2007 (UTC)

Maybe I'm missing what you are trying achieve. If you want to know the position of a vehicle at all times given engine power curve and gearing, it can be done. It's hard to develop a closed form solution as engine power curves are not easy to model (unlike some types of electric motors). If you can come up with a reasonable engine model, I can derive the equations of motion, differentiate once for velocity and once more for acceleration. We can even differentiate again and find the point of maximum acceleration if you like. Of course, this assumes all variables are known, something that is not completely clear to me in your statements above. Madhu 14:43, 17 March 2007 (UTC)

All good questions, Madhu. Yes, I'm being pedantic but, I'm of the firm opinion that most of the arguments in the world occur because of the use of imprecise language. Maybe we can talk about that later on your talk page.

This whole debate began when the following was inserted into the main article:

In the automotive enthusiast community it is a widely held misconception that the torque rating of a motor is an indication of how rapidly it will be able to accelerate a vehicle. Acceleration is work, and as such requires the application of force over distance (the definition of power). Horsepower or watts provide the indication of the power-producing capability of a motor.

I removed this text and stated that it was obviously incorrect. Neglecting the obvious errors (a = w, and P = Fd), I said that the acceleration is proportional to torque, not power. Since, by acceleration, I mean (dv/dt), my statement is an undisputed fact. However, the author of the orginal statement is apparently using the phrase rapid acceleration to mean largest average acceleration (correct me if I'm wrong here, IBJ). So, a massive misunderstanding ensued. Thus, my effort here is to simply try and find the answer to what the largest average acceleration is in fact proportional too. And, according to my reasoning, the largest average acceleration is (in general) not determined by max torque alone, or max power alone, but is determined by power bandwidth. As you said, in the limiting case of a continuously variable transmission, the answer is easy. The engine with the highest power output provides the largest average acceleration. However, in the case of a discrete transmission with a finite number of gear ratios, it seems possible for an engine with less maximum power to provide larger average acceleration than an engine with higher maximum power rating.

So, the statement that engine power output determines average acceleration is not generally correct either.

Regarding modeling, I've been giving that a lot of (too much, in fact!) thought lately and here's what I think the results would look like when I find the time to do it in MATLAB.

First, as I shown above in a much earlier post, the velocity has a t^1/2 dependence in the case of a continously variable transmission (CVT). This is the upper bound. In the case of a discrete set of gears, start with a constant torque engine. Then, v is proportional to t so we'll get a piecewise linear v over t curve that is always under the CVT curve and sets a lower bound. Now, look at what happens if you add a torque peak keeping the max engine power constant and the shift points constant. We get a velocity curve that sits in between. Thus, we can say that increasing the max torque of the engine without changing the max power output increases the average acceleration. Alfred Centauri 15:37, 17 March 2007 (UTC)

If you are saying a good transmission can make a lower power engine look better than a higher power engine with a lame transmission, then yes, I agree. But, it sends the wrong message. It's always possible to make a high power engine look bad. In the case of commercially available engines, we don't have much choice in the distribution of power over it's operating range. We do have a choice of how much peak power it produces. In the case of transmissions, we have a wide array of choices. For economic reasons, auto manufacturers choose to use sub-optimal drive trains. With increasing fuel costs (among other factors), this is changing and there are cars available today with better drive trains that use engine power more efficiently. Madhu 16:46, 17 March 2007 (UTC)

Racing associations impose arbitrary restrictions on engines, fuel, intake, and drive trains. They want to level the playing field because they know a well funded team could leave everyone else in the dust with a few changes to the drive train. One obvious, commercially available solution is a high power/weight ratio engine driving a generator driving electric motors on each wheel. This is not rocket science. As you know, it's at least 50 (if not 100) year old technology. Of course, it all starts with a high power/weight ratio engine. If you want to start with a low power/weight ratio engine and fool around with a handful of fixed gear ratios, go for it. But don't be surprised when the next generation hybrid economy car smokes the last generation five speed muscle car at stop light. Madhu 16:46, 17 March 2007 (UTC)

I'm comparing engines using the same transmission so your point is not the point I'm working towards. Same transmission (not CVT), two engines. One engine produces less max power but produces that horsepower over a wide range of rpm. The other engine produces higher max power but only over a narrow range of rpm. Alfred Centauri 17:43, 17 March 2007 (UTC)

[edit] Does acceleration depend on power?

Is the statement "[the] acceleration of a moving object is dependent on the power" a generally true statement? To be more specific, does a = P/mv always?

A vehicle climbs a hill at a constant speed. The kinetic energy of the vehicle does not change as it climbs the hill since the speed is constant. At all times, the engine is developing power and some of that power is delivered to the wheels.

Does a = P/mv?. The vehicle is moving. The power is not zero, the velocity is not zero, so the acceleration is not zero. But, this is a contradiction. The acceleration must be zero because the vehicle maintains a constant speed.

So, a = P/mv is not true in general. Since there is power delivered to the wheels, the energy of the vehicle must be increasing if energy is to be conserved. The kinetic energy is not increasing since the vehicle speed is constant. What energy is increasing?

The answer is, of course, the potential energy: U = mgh. The power delivered to the wheels from the engine increases the potential energy of the vehicle.

In general, the energy of the vehicle is:

$E = \frac{1}{2}mv^2 + mgh$

This gives:

$P = mv[a + gsin(\theta)]\$

where $θ$ is the angle of the velocity from the horizontal.

A more generally correct statement is then:

$a = \frac{P}{mv} - gsin(\theta) = \frac{F}{m}$

a = F/m will always work. Alfred Centauri 20:37, 17 March 2007 (UTC)

[edit] Does the force at a non-zero velocity come from the power?

The force at a nonzero velocity comes from the power. Please try to understand that. And yes, a vehicle going up a hill will lose some power to PE, but the rest will go to KE. Point still stands. 74.101.94.34 05:18, 18 March 2007 (UTC)

Thank you for being so patient with me. However, power is not lost to PE any more than it is lost to KE. In this case, the power is simply equal to the rate of change of PE.

How is power lost? Power is lost through a mechanism like friction. But PE can be converted into KE so no power is lost here.

In the example above, there is power delivered to the wheels but the force (torque) at the wheels is zero. This directly contradicts what you said: "The force at a nonzero velocity comes from the power".

But, more importantly, you've got it backwards. It's easy to see this.

Apply a constant force to an object that is not subject to any other forces and that is initially at rest. Let's say the force comes from a rocket producing constant thrust. The object accelerates at a constant rate: constant force <==> constant acceleration (well, this is true if the total change in object mass is neglible).

Look at the power though. The power starts at zero and steadily increases with the speed of the object.

So, how does the rocket engine know to produce more power? Don't you see? P = Fv in this case. It is a derived quantity. It is the force developed by the rocket engine that is fundamental.

Here's a different perspective: Two people measure the power and force developed by the rocket at the exact same instant. They agree on the force measured but disagree on the power measured. Why? The people are in (constant) relative motion with respect to each other, so they measure a different speed for the rocket. P = Fv. Same force, different relative velocity, different power.

If power is fundamental and if force is due to power, the measured acceleration of the object would depend on the (constant) state of motion of the observer But, this is not seen in nature. At least in the realm of non-relativistic physics, acceleration is absolute. All observers in constant motion measure the same acceleration. This is a well known result. So, force is absolute, not power!

Don't you see this? Alfred Centauri 13:29, 18 March 2007 (UTC)

Retrieved from "http://en.wikipedia.org../../../h/o/r/Talk%7EHorsepower_9786.html"

Talk:Horsepower

From Wikipedia, the free encyclopedia

Contents

[edit] Now

[edit] Real power of the horse?

[edit] Original horse power

[edit] Derivation for HP Equation

[edit] Pound mass

[edit] Determining horsepower

[edit] Horse-power machine

[edit] Morven's comments from User talk:Bobblewik

[edit] Coping with ambiguity in the term 'horsepower'

[edit] Ambiguity in PS

[edit] The Most Powerful Diesel Engine in the World

[edit] POV

[edit] hp (ECE)

[edit] Removed Link

[edit] Metric Units

[edit] Removal of text

[edit] Removal of text (reprise)

[edit] Added disputed tag

[edit] Torque, power & acceleration: Two misunderstandings

[edit] New info on hp versus torque debate

[edit] Unnecessary Debate

[edit] A suggestion in the hopes of removing the dispute tag

[edit] The penny drops, acceleration, power and torque

[edit] Two impossible engines

[edit] A choice of two engines

[edit] Does acceleration depend on power?

[edit] Does the force at a non-zero velocity come from the power?

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