Wikipedia:Reference desk archive/Mathematics/2006 September 1

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Contents 1 conversion 2 3D model from 2D pictures 3 roman numerals 4 Finance: Hagan volatility

[edit] conversion

calculate the molarity of the following solutions. 80g of NaOH in 500mL of solution. thanks Johanna

Do your own homework: if you need help with a specific part or concept of your homework, feel free to ask, but please do not post entire homework questions and expect us to give you the answers. Letting someone else do your homework makes you learn nothing in the process, nor does it allow us Wikipedians to fulfill our mission of ensuring that every person on Earth, such as you, has access to the total sum of human knowledge. Also, Wikipedia:Reference desk/Science is a more appropriate place for any question you may have about this problem. -- Meni Rosenfeld (talk) 12:27, 1 September 2006 (UTC)

Yes, it does sound like homework, but here are some hints to get you started. First, work out the molecular mass of NaOH. From this you can calculate the mass of 1 mole of NaOH. Then work out how many moles there are in 80g of NaOH. Now, you have this many moles in half a litre - so how many moles are there in 1 litre ? This is the molarity of your solution in mol/litre. Gandalf61 13:01, 1 September 2006 (UTC)

[edit] 3D model from 2D pictures

The picture on the right is of an actuator that can move in three dimensions. Assume the actuator to be long flexible cylinder. I want to get approximate equations describing the centerline (something like {x(s),y(s),z(s)}, where s is the arclength) of this cylinder from photos taken from different angles. I can take as many photos from any position/orientation. How can I do this? If someone knows a way to create a 3D rendering (that can be rotated around) from 2D pictures, that would also be awesome (I'll still need to get the equation of the centerline.) Thanks a lot! deeptrivia (talk) 15:46, 1 September 2006 (UTC)

I don't know how to do it, but this sounds similar to Photogrammery. You might look up that topic. I couldn't finda a Wikipedia article on Photogrammery. --Gerry Ashton 16:39, 1 September 2006 (UTC)

Oh, I guess it's the same as Photogrammetry. Let me add that I don't nececssarily need a full fledged 3D model. Possibly finding the coordinates of a few dozen predefined points in 3D will suffice. deeptrivia (talk) 16:41, 1 September 2006 (UTC)

If you take two pics from right angles, say the top view and the side view, that should be enough to identify the centerline in 3D. You can put a grid over each pic, then label points that cross the grid:

Y
^
|      TOP VIEW
 +-----------------+
6|                 |
5|                 |
4|      **         |
3|   ***  ***   ** |
2|           ***   |
1|                 |
0+-----------------+
 0 1 2 3 4 5 6 7 8 9 -> X

Z
^
|     SIDE VIEW
 +-----------------+
6|                 |
5|        ******** |
4|      **         |
3|     *           |
2|    *            |
1|   *             |
0+-----------------+
 0 1 2 3 4 5 6 7 8 9 -> X

In this example the first point has coords (2,3,1) and the last point has coords (8,3,5). Note that the pics should be taken from a far away as possible, with a zoom lens, to minimize perspective view distortion. Ideally, we would want perfect orthographic views. StuRat 21:15, 1 September 2006 (UTC)

As for creating a 3D model, most 3D CAD systems will run a curve through a series of 3D pt coords. It may take some trial and error, though, to avoid distortions of the curve. You might try different degrees, constraints, and curve types (Bezier, for example), to get the most accurate curve. Surprisingly, more sample points doesn't always produce a better curve (more pts can make for a lumpier curve). Which 3D CAD systems do you have at your disposal ? See curve fitting for a discussion. StuRat 21:28, 1 September 2006 (UTC)

Thanks. I won't be using CAD. I've got to do all the stuff in MATLAB, but I think curve-fitting won't be an issue. Do you have any practical tips on how to place the cameras to that they take pictures from two side views that are almost perfectly orthogonal, and the photos from them are of the same scale? I appreciate your help. deeptrivia (talk) 20:29, 2 September 2006 (UTC)

Just the suggestion of taking them from the maximum possible distance which will still allow the object to fill the frame at the maximum possible zoom. As for getting the scale to be the same, don't worry about that. As long as the aspect ration is the same, you can just scale one up digitally on computer so the two lengths will match. Note that you can either move the camera or rotate the object by 90 degrees, whichever is easiest. Also, you want to pick two orthogonal views which show the most detail, or longest length, in this case. For example, if you had an approximately linear object, you wouldn't want an end view, where it would look almost like a point. StuRat 23:08, 2 September 2006 (UTC)

You can take an picture of it, then move some measured distance to the left and take another picture. From these two pictures you can examine the horizontal displacement between the same object in both pictures and use that to construct a depth map of the shape as viewed from that side. If you did this a few times from different angles, you could combine the data to try and reconstruct the visible surface (completely hidden surfaces would be impossible to reconstruct from the pictures though). Also, Stanford has an interesting site about constructing 3D models using high quality 3D scanners. - Rainwarrior 04:45, 3 September 2006 (UTC)

You could do that, but that requires use trigonometry to calculate each point's coords. Taking pix from right angles is far simpler, and, as a bonus, the error is less. StuRat 07:56, 4 September 2006 (UTC)

By "different angles", I meant right angles, more or less, but really just whatever angles you need to cover the visible surface. Simply taking pictures from right angles and using a convolution of their silhouettes is a way simpler operation, but will have WAY more error (consider a surface at a 45 degree angle to your view... it will show up as a box). The method I am suggesting provides a depth map for every view which would pick up that error, and as well could be applied to other angles as long as the location and direction of the camera is known. - Rainwarrior 18:31, 4 September 2006 (UTC)

P.S. The only source of inaccuracy would be from measuring the camera position, as well as uniform surface colourings (which disguise depth) and changing specular highlights (if the surface is shiny). You could improve accuracy significantly by painting with a very diffuse (rough) paint in different coloured patches to improve point matching. Computers are quite capable of doing accurate trigonometric calculations (that's what they're made for), so that isn't really a source of error. - Rainwarrior 18:36, 4 September 2006 (UTC)

This is one of the key problems in computer vision I've been to conferences where a large proportion of the papers were relating to this problem. The big problem is in matching points, if you can project a grid over the surface or identify specific points by hand things get easier. I seem to recal that there are also problems with the numeric stability, small inacuracies in measurements can lead to bigger errors in the calculated position. The general solution to these errors is too use multiple images and combine the data from those, this then becomes a tricky statistical problem as to how to average the different points correctly.

If you can't lable the points by hand this then becomes a very interesting problem and there are all sorts of sosphiticated algorithms, in the trade called Scene reconstruction (alas no article). Motion can often play an important part as between one frame and the next the points will only move a small amount.

Some 3D scanners use a different technique to just taking photos, I think they use time of flight of lazers pluses to measure distance. --Salix alba (talk) 19:15, 4 September 2006 (UTC)

In my experience with the technique, small errors in the computer's calculations don't make any impact on the results (maybe there are other techniques for this I don't know about), but errors in the human measurement of the camera can make a lot of difference when the camera is up close. But, if you take enough measurements from different angles in a higher resolution than you need, smooth out the results (and filter errors) and combine them (simplify the surface), it can work out quite nicely.

To my understanding 3D scanners don't take photographs at all, really (because of all the problems of matching). It's more accurate just to take depth measurements at a fine resolution. Colouring is a problem that can be solved seperately from the geometry and applied later. (But with the matching techniques, you can do them both at once.) - Rainwarrior 19:36, 4 September 2006 (UTC)

Oh, also, taking pictures from many angles and using only the silhouette, you can construct a pretty good shape of an object as long as it's convex. (It's simpler than doing all that pattern matching, at least.) - Rainwarrior 19:46, 4 September 2006 (UTC)

If you just want the centre line, though, you could probably do a pretty good job with just two photographs from different angles... the projection doesn't actually need to be orthographic, either, as long as you know where your cameras are. If you can pick out some critical points by hand on the pictures (or come up with some recognition scheme) you should be able to find the points in 3D as the intersection of two rays, one from each camera to those points. - Rainwarrior 19:51, 4 September 2006 (UTC)

Everyone else here seems to assume he has the appropriate software to perform photogrammetry. I did not, so showed him how to do this manually. StuRat 04:26, 6 September 2006 (UTC)

Well, when I was doing work on these problems I was using MATLAB, which he says he is already using. (And if he has any questions specifically about how to implement these ideas in MATLAB I would be happy to discuss it.) - Rainwarrior 07:48, 6 September 2006 (UTC)

[edit] roman numerals

what year, when written in roman numberals, uniquely contains one each of the Roman number symbols in descending order?

Go look at out Roman numeral article and figure it for yourself. StuRat 20:55, 1 September 2006 (UTC)

Either MDCLXVI (1666) or MDCLXVMDCLXVI

(1,666,666). dpotter 22:18, 1 September 2006 (UTC)

Or 1,666,666,666 if you use double overlines or underlines to extend Roman numerals further: MDCLXVMDCLXVMDCLXVI. StuRat 01:02, 2 September 2006 (UTC)

[edit] Finance: Hagan volatility

Does anyone know what a hagan or haganised volatility is in Financial risk terminology, how it can be computed or what data is required to compute it?

'Hagan' volatility is a formalisation of intuitive ideas about volatility. I can't help thinking that this question and the preceeding one were posted by people under the impression that volatility is some inherent property of an asset which will enable a speculator to predict its price in the future - it isn't.

As to using mathematical formulae to gain a trading edge, in general it's a pipe dream for reasons that are far too complicated for me to go into at 1AM.

If you're determined to pursue this sort of thing then my advice is that you should be looking at alternative option pricing strategies, specifically ones that use historical price distributions rather than inaccurate assumptions about normal price distributions - this is something with proven efficiences vs. the market and which can be calculated with a PC rather than a cluster. Rentwa 00:32, 2 September 2006 (UTC)

Can anyone answer my question without waffling?

The answer to your question is: 'If you have to come to Wikipedia to ask about volatility then you haven't a hope in hell of understanding financial mathematics or making any money trading.'

Clear enough for you? Rentwa 11:17, 2 September 2006 (UTC)

Perhaps this article answers your question. --Lambiam^Talk 15:23, 2 September 2006 (UTC)

Thank you, Lambian. That is an excellent resource. In answer to Rentwa - stop trying to second-guess people's reasons for seeking knowledge, don't judge everyone by your own poor standards.

Then don't be rude to people who are trying to help you. If you weren't happy with my answer then you could have said so politely. If I've misjudged you then I'm sorry. I've answered hundreds of questions about financial maths in my years on the internet and in my experience most people asking them have come across some arcane sounding phrase they don't understand but think will lead to riches, and in almost every case they believe this because they don't have any grounding in maths or finance.

'Hagan' volatility is highly convoluted, doesn't tell you very much and doesn't feature in any realistic trading strategy I'm aware of. Out of the goodness of my heart I tried to point you to a workable, relatively new (and therefore still profitable), reasonably simple strategy you can employ using commercially available software on a PC, but still managed fall short of your standards! Once again, I apologise. I hope that one day I may elevate myself to a level from where I can at least appreciate you, and maybe even learn from you. Until that day, I remain yours, Rentwa, slithering around in the mire.