Talk:Trajectory

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what itrajectory in a simpler explanation

The opening paragraph is a pretty simple explanation: the path of an object through space. What sort of explanation are you looking for? Samw 23:47, 1 October 2005 (UTC)

"To neglect the action of the atmosphere, in shaping a trajectory, would (at best) have been considered a futile hypothesis by practical minded investigators, all through the Middle Ages in Europe."

What's this about? Practical-tminded investigation was not the hallmark of the academics of the Middle Ages and early Renaissance. I would have been nice if they'd written mathematical treatises on, for instance, the proper construction of cathedrals (and it would in principle have been possible from the 13th century on), but they didn't; they left that grubby stuff to the trial and error of engineers. (Three guesses who did write the first treatise on mathematical principles of structures. Hint: he's mentioned in the article.) It's true that they wouldn't consider what happens in a vacuum -- because they knew that Nature abnors a vacuum (Aristotle says so); also, the idea of a vacuum is arguably contrary to religion. We could put in that information to mock the poor old Middle Ages, but I think it's more respectful just to zap the passage entirely. -- Dandrake 08:04, 12 December 2005 (UTC)

You're encouraged to re-write this verbiage and other verbiage in the article. While I added most of the equations in the article, the introductory text leaves something to be desired. Samw 10:09, 12 December 2005 (UTC)

Contents

[edit] Derivation of Rifleman's rule

I received an email flagging a possible problem in the derivation section of "Uphill/downhill in uniform gravity in a vacuum":

Hi Sam,
I was using your trajectory page(nice work) and I am having a problem with part of the derivation. Look at the equation that is fourth from the bottom of the page (this equation is correct). The factoring of sin(Theta)/cost(Theta) appears to be done improperly in going to the next equation.
Could you check this? If this is true, it would affect your comments on this equation that were made earlier in the article around Equation 11.
Mark

My response:

Hmm, you seem to be correct.
The derivation was my work but equation 11 was taken from equation 1-29 at [1]
And certainly the equation for "Rifleman's rule" is an established concept. But you're right, the derivation seems to be incorrect.
Maybe over the holidays I'll have time to figure it out &/or look up the referenced text. In the meanwhile, I'm going to post this errata on Wikipedia and see if others can help. Thanks for contacting me on this and don't hesitate to update the page yourself!
Sam

If anyone can help sort this out, that would be greatly appreciated! Samw 04:52, 25 December 2005 (UTC)

Hi Sam,

I have been looking at some references and I think I may know what is going on here. The original reference is known to have many typos (see the book reviews at [2]).

I believe the equation in error should have a cotangent instead of tangent in one of its terms.

R_s=R(1-\cot\theta\tan\alpha)\sec\alpha \;

I have simulated this situation using an ODE solver and have verified that this equation works and the one in the article does not.

I think I have been able to derive the "Rifleman's Rule" from this expression. It takes a bit of work.

Assume that a rifleman has sighted in his weapon on a flat surface at some range we can call R0. Because the bullet travels along a parabola, the bore of the weapon will have an angle with respect to the line of sight (LOS). Let's call this angle dθ. We can compute this angle using the following equation.

R_0=v^2\sin2d\theta/g \approx\ v^2 2d\theta/g for small dθ.

When the rifleman attempts to shoot uphill with his weapon zeroed at R0, the gun will actually shoot further than expected (as we will see). Assume that when shooting uphill (hill angle =α), the riflebore is at an angle of α + dθ with respect to gravity. We can write an expression for the bullet point of impact up the hill as follows.

R_s= \frac{v^2\sin(2(d\theta+\alpha))}{g}(1-\cot(\alpha + d\theta)\tan\alpha)\sec\alpha \;

We can approximate 1 − cot(α + dθ)tanα)secα using a Taylor approximation as shown below.

R_s= \frac{v^2 2sin(\alpha)cos(\alpha)} {g}\frac{sec^2(\alpha)}{tan^2(\alpha)}d\theta sec(\alpha)

After some simplification, we obtain the following expression.

Rs = R0sec(α)

Because the sec(\alpha)\ge 1, we see that the bullet will hit the hill a bit further up than where the weapon was zeroed. This range extension can be eliminated by firing the weapon as if it were being aimed at shorter horizontal distance Rcorrection = R0cos(α). The following equation illustrates this point.

Rs = R0sec(α)cos(α) = R0

Does this seem reasonable? Note that the "Rifleman's rule" is an approximation but probably holds well for typical situations.

Mark

Wow, quick analysis. Errata for "Modern Exterior Ballistics" is listed at [3] but it doesn't list the cot/tan mixup. Then again, it's not clear that 1-29 came from this book; I only used the webpage I gave and don't actually have the book. Yes, my own derivation clearly points to cot not tan, so I can't object to you replacing equation 11 in the article with the cot version. Furthemore the immediate discussion afterwards on downhill firing and critical angles for uphill firing applies (though the critical angle equations will have to be adjusted.)
As for derivation of the Rifleman's rule, I unfortunately couldn't follow the "expression for the bullet point of impact up the hill as follows" (though I admit I didn't try very hard). May I suggest a step by step explanation? Since my derivation is unfortunately clearly wrong, I'm fine with you replacing my explanation right in the article.
As a sanity check on the whole affair, I must admit I'm uncomfortable on how a tan/cot mixup in equation 11 can still lead to the same Rifleman's rule. I'm half hoping for a second error in my original explanation of Rifleman's rule so this all makes sense again! Samw 21:27, 26 December 2005 (UTC)

Hi Sam,

I agree with you that a step-by-step development is required. I am new to the Wikipedia, so I am still learning how to do things, like adding a figure (which this derivation desperately needs). Also, I really should look at the original reference.

Thanks for the errata reference. With a little more time, I will try to put something more complete together.

For now, I have been able to solve my immediate problem with the ODE solver. However, an equation-based solution would be more useful to the general public. blacksheep 00:57, 27 December 2005 (UTC)

For images and Wikipedia, see Wikipedia:How to edit a page#Images. I'd be happy to help in anyway I can. Samw 01:57, 27 December 2005 (UTC)

Hi Sam,

I have found the key issue. The definition of θ in the referenced web article ([4]) is different than the definition of θ in your work. With the change of variables and a massive amounts of trigonometric simplification, I have been able to show that your work (minus the one error) and the reference work are equivalent.

I prefer your derivation to the reference article (it uses a coordinate rotation that complicates the algebra). However, the derivation in the reference article does allow a simpler path to the "Rifleman's Rule." blacksheep

Hmm, so in order to derive Rifleman's rule, we have the choice of "massive amounts of trigonmetric simplication" to get to my version of equation 11 first or a complex derviation using Taylor's series expansion? Sigh. Now I know why I've never seen a good derivation of Rifleman's rule.  :-) Thanks for the tidbit. I'll adjust the accuracy-disputed flag in the main article to cover equation 11 as well. There's no rush on resolving this; it's sat like this for months! Thanks. Samw 18:42, 28 December 2005 (UTC)


Hum, looking through this page there seems to be entirely to much mathematics. Do we really need derivations of formula in an encyclopedia? It might be better just to show key results. See Wikipedia:Manual of Style (mathematics)#Proofs and some of this might be bordering on Original research. Its also giving a bit too much away to students who have this set as home work! --Pfafrich 22:37, 2 January 2006 (UTC)

I'm flattered you think this might be original research! I assume the concern is with Rifleman's rule itself; not the basic trajectory equations. For Rifleman's rule, I have no objections to this material moving to a separate article on Rifleman's rule. Let me know if that's your opinion, and I can do that. There are enough references that clearly this isn't original research. Thanks to the MoS on proofs; I'll try to isolate the derivations more. However, I think the basic equations and derivations are common enough that it should be included. If it's likely to be assigned as homework, all the more reason that this is common enough to be included! Note that there's still trajectory at an elevation and other variations that are probably beyond the scope of Wikipedia.
Now back to the actual issue that triggered this discussion; any thoughts on the actual derivation of Rifleman's rule? Thanks. Samw 04:01, 3 January 2006 (UTC)


Its still way to complicated. Heres a simpiler derivation

We know solution is a parabola and hence if we write p for the position and t for the time we have

p = (At,0,at2 + bt + c)

the first and second deriv is

p' = (A,0,2at + b)
p'' = (0,0,2a)

At t = 0 we have

p=0,\ p'=(v_h,0,v_v),\ p''=(0,0,-g)

hence

A = v_h,\ a = -g/2,\ b = v_v,\ c = 0.

Giving eqn of parabola as

p = (v_h t,0,-0.5 g t^2 + v_v t)\,.

To find the (horizontal) range need to find value of t where z-componant of p is zero i.e.

-0.5 g t^2 + v_v t = (-0.5 g t + v_v) t = 0\,

hence soln are at t = 0 and t = 2vv / g subs back in eqn for x component of p gives

R = 2 v_v v_h / g\,.

By symetry max height will occur when t = vv / g giving

p=(v_v v_h / g, 0,0.5 v_v^2 / g)\,

To find derivation in terms of angle of inclination θ and speed of projectile v we have

v_h = v \cos(\theta),\ v_v = v \sin(\theta)

substituting into eqn for range this gives

R= 2 v^2 \cos(\theta) \sin(\theta) / g = v^2 \sin(2\theta) / g\,.

Note that calling this polar coordinates is not strictly correct, we still use cartesian coords but just express initial conditions in terms of speed and angle.

Really thats all which is needed for entire derivation, the rest can easily be dropped.

To find the rifle mans range we wish to find the intersection of p with the curve

q = (v_h t , 0 , v_h t \tan(\alpha))=t v_h (1,0,\tan(\alpha))\,

where α is the angle of the hill. (I've parametrised the line by t to make things simpler). Equating the z-components and dividing through by t (OK as t=0 is starting soln)

v_h \tan(\alpha) = -0.5 g t + v_v\,

hence

t= {2\over g}(v_v - v_h \tan(\alpha))\,

this gives

p={2 v_h \over g} (v_v-v_h \tan(\alpha)) ( 1, 0 , \tan(\alpha))

the length of this is

R_s={2 v_h \over g} (v_v-v_h \tan(\alpha)) \sqrt{1+\tan(\alpha)^2}

using the trig identity 1 + tan(α)2 = sec(α)2 gives

R_s={2 v_h \over g} (v_v-v_h \tan(\alpha)) \sec(\alpha)

Divide by R={2 v_h v_v \over g} gives

{R_s \over R} = (1-{v_h\over v_v}\tan(\alpha))\sec(\alpha)

Finally

{v_h\over v_v} = {v \cos(\theta)\over v \sin(\theta)}=\cot(\theta)

so

{R_s \over R} = (1-\cot(\theta)\tan(\alpha))\sec(\alpha)

So I agree its cot not tan. --Pfafrich 14:06, 3 January 2006 (UTC)

Sorry I wasn't clear on my question. Yes, User:Ziggle and I came to the same conclusion on the range for uphill/downhill; that equation 11 in article should contain a cot and not a tan. The outstanding issue is how to derive "Rifleman's rule" given the cot. Given all the references on Rifleman's rule, I'm pretty sure the rule itself is valid. Since equation 11 contains a cot, the explanation in the article is no longer valid. Any suggestions on how to derive Rifleman's rule given the revised equation 11?
I see your point now about the overkill in the various derivations. I'll try to find a home for all this in Wikibooks. Samw 22:20, 3 January 2006 (UTC)


Its a bit late at night for me, so apoligies if this is giberish. Still trying to workout what riflemans rule is all about, (I know some maths but nothing about shooting). One point did occur is that the a good aproximation is \sin(\theta) \approx \theta when angle is given in radians. This is a bit more acurate than just saying its zero. This may help some.

I'll have a bash at tiding up the main article tomorrow. Yep it does seem like wikibook would be a good home. --Pfafrich 00:22, 4 January 2006 (UTC)

Hi folks,

When Sam and I were discussing this page, I did put together a derivation of the rifleman's rule based on this cotangent version. I have not updated the article because I was concerned that it was a bit much. I have put together a web page that summarizes my work([5]). I am sure it could be simplified, but it reflects how my head was working.

blacksheep 22:49, 5 January 2006 (UTC)

I like it, I certainly some pictures would go well in the main article. A little quible with using dφ d normally stands for infinitesimal derivative. Here you really using it as a difference δθ might be more appropriate, as δ is often used for differences.

I've not looked through the maths yet. I'll give it a ponder.

Maybe it time for a new page Riflemans rule, it seems excessive for trajectories. Possible wikibooks? --Pfafrich 00:49, 6 January 2006 (UTC)

Hi folks,

I have never written a Wikibook, but would like to learn. The "rifleman's rule" does seem to be a well known rule that is missing a good, generally available derivation.

blacksheep 04:35, 6 January 2006 (UTC)

User:Pfafrich is proposing a Wikipedia article. Just click on the red link Rifleman's rule when you're signed in and start typing! Unless the article is overwhelming in its detail, I don't think it needs to be moved into Wikibooks. Wikibooks is for materials deemed outside the scope of a general encyclopedia. Hopefully tomorrow I'll have time to move the attached multiple derivations for general trajectories over to Wikibooks. I was thinking of adding it to [6] and cross-referencing it here. Samw 04:57, 6 January 2006 (UTC)
I've added a reference to Wikibooks in the main article. Samw 15:34, 6 January 2006 (UTC)

[edit] New page Rifleman's rule

Theres a reciently created page Rifleman's rule wondering if we should trim this page a bit, to save repeated material. --Salix alba (talk) 14:32, 31 March 2006 (UTC) (was PfafRich)

Trim this talk page or the trajectory article? The incline fire section of the article needs fixing. Length of the talk page is harmless. Samw 01:06, 1 April 2006 (UTC)

[edit] Link I'd like to add

Hello folks I would like to add this link to the article: Projectile Lab. It's a JavaScript based trajectory simulator I wrote that asks the user to calculate aspects of the projectile. Any objections? — Edward Z. Yang(Talk) 21:30, 28 December 2006 (UTC)

Added .— Edward Z. Yang(Talk) 21:38, 10 January 2007 (UTC)

[edit] Almost incomprehensible

Yow. This page is nearly incomprehensible. Consider doing a rewrite using more English and less math.

The math itself is not consistent. Equation 11 shows cotangent of theta; all the following equations use tangent theta. I assume that cotangent theta is correct, since in the limiting case alpha=theta, the range should be zero. Thus, it looks like everything after equation 11 is wrong.

(looking up at the talk before my post, it looks like you already understand this, but did not correct the page)

I think that a difficulty is that you have never defined whether theta is measured relative to the horizontal, or relative to the surface. I will presume that it is defined relative to the horizontal.

I'm going to delete all the sections that have tan theta instead of cot theta, which is the part between equation 11 and the rifleman's rule. I'm leaving in the rifleman's rule, since I assume it's correct, even though it's not clear how it comes from equation 11.

I find this sentence very difficult to follow: Thus if the shooter attempts to hit the level distance R, s/he will actually hit the slant target. "In other words, pretend that the inclined target is at a horizontal distance equal to the slant range distance multiplied by the cosine of the inclination angle, and aim as if the target were really at that horizontal position."

What in the world are you saying here? The article is about how far a projectile travels. Geoffrey.landis 03:32, 19 March 2007 (UTC)

Yes the uphill/downhill section has problems, hence the "disputed" tag. In the meanwhile, can you revert back to the orignal form of equation 11? The Rifleman's rule article keys off that. By all means re-write; I haven't had the skill (or time) to correct the error in the derivation. Samw 03:28, 20 March 2007 (UTC)
As originally written, equation 11 was clearly, unambiguously wrong. Why exactly do we want to revert to a "original form" that is wrong? Not having it at all would be better than having it wrong, I would think. If the following material "keys off of" material that is wrong, rather than replacing the right material with wrong material, perhaps we should delete it-- it is only marginally relevant to the discussion of trajectories anyway. (Another suggestion would be to move all the Rifleman's rule discussion from here to the Rifleman's rule page, and just link that article). Geoffrey.landis 14:04, 29 March 2007 (UTC)
I think the derivation is wrong. Equation 11 is correct per the discussion above. Hence the disputed tag. Samw 22:56, 29 March 2007 (UTC)
If equation 11 is correct (as you say) then the following equation (which is unnumbered) must be wrong since it has a tangent in the place where equation 11 has a cotangent. So, since we both agree that it is wrong, why do you want to restore it? Again, I would say that it would be better to have no information than to have wrong information. Geoffrey.landis 02:32, 12 April 2007 (UTC)

[edit] Angle of Elevation section addition

Can we consider adding the following to the "Angle of elevation" section? I don't know how exactly to fit it in, but it is an equation solving for angle with differing initial and final heights. g is acceleration of gravity (e.g. 9.8m/s), v is initial velocity, R is desired range, and Delta h is final height minus initial height. It is self-derived and tested, although I am sure someone more important has derived this before. Unfortunately, it would be almost impossible to find an equation like this on google.Geogriffin 23:33, 29 April 2007 (UTC)

\theta = \arctan({-R \pm \sqrt{R^2 - 4A(A-\Delta h)}\over 2A}) where A = {-gR^2\over 2V^2}
Go for it! Just clearly define the terms as I'm not sure what you mean given the terse description above. Samw 03:14, 30 April 2007 (UTC)

[edit] Anticipating the existence of the vacuum?

...by anticipating the existence of the vacuum...

Calculating as if the object moves through a vacuum is not the same as anticipating the existance thereof. Rather, what is done is ignoring the friction from the air (drag), since the contribution thereof is often small and the formulae are a lot easier without it. Shinobu 11:04, 12 May 2007 (UTC)

Be bold and by all means reword that section. Samw 13:23, 12 May 2007 (UTC)