Lee Mac Posted May 18, 2009 Share Posted May 18, 2009 Ok, so now that we have the method for finding the Maximum and Minimum points on the curve, how can this be applied to find the signs of the vectors? Quote Link to comment Share on other sites More sharing options...
lizp Posted May 18, 2009 Author Share Posted May 18, 2009 Sorry for not being able to post yesterday. Will be busy today too but will try to lurk from time to time as much as I can. Now that we have the maximum and the minimum of the curve, all base vectors that are applied to points of the curve that are to the left of both its maximum and minimum will lead to positive torques while all base vectors to the right of both these extremums will lead to negative torques. Quote Link to comment Share on other sites More sharing options...
Lee Mac Posted May 18, 2009 Share Posted May 18, 2009 Sorry for not being able to post yesterday. Will be busy today too but will try to lurk from time to time as much as I can. Now that we have the maximum and the minimum of the curve, all base vectors that are applied to points of the curve that are to the left of both its maximum and minimum will lead to positive torques while all base vectors to the right of both these extremums will lead to negative torques. OK, I understand: Min Max > Min > x = negative torque But what about Max > x > Min ? Quote Link to comment Share on other sites More sharing options...
lizp Posted May 18, 2009 Author Share Posted May 18, 2009 OK, I understand: Min Max > Min > x = negative torque But what about Max > x > Min ? Good question. Could it be done like this: all points that are to the left of Max and above y-axis of the center and to the left of Min and below y-axis of the center are positive. All the rest are negative. Quote Link to comment Share on other sites More sharing options...
Lee Mac Posted May 18, 2009 Share Posted May 18, 2009 OK, I shall see what can be done Quote Link to comment Share on other sites More sharing options...
Lee Mac Posted May 18, 2009 Share Posted May 18, 2009 OK, the signs should be correct on this one (hopefully!) Vector Analysis.lsp Quote Link to comment Share on other sites More sharing options...
lizp Posted May 18, 2009 Author Share Posted May 18, 2009 I'm getting an error message: 'Extremities Could No be Ascertained' Quote Link to comment Share on other sites More sharing options...
lizp Posted May 18, 2009 Author Share Posted May 18, 2009 Oh, I see now. I'm checking your last code on the original drawing -- that with the "curve" made up of numerous small segments. That's why I'm getting the error message. When I checked the code with your drawing of the track as a contiguous curve the values are displayed and their sign seems to be correct, except for the sense of the torque-generating vectors (Cyan) in the upper left quadrant. Quote Link to comment Share on other sites More sharing options...
Lee Mac Posted May 18, 2009 Share Posted May 18, 2009 Oh, I see now. I'm checking your last code on the original drawing -- that with the "curve" made up of numerous small segments. That's why I'm getting the error message. When I checked the code with your drawing of the track as a contiguous curve the values are displayed and their sign seems to be correct, except for the sense of the torque-generating vectors (Cyan) in the upper left quadrant. Oh yes, I added that message so that you would know when the Maximum and Minimum points could not be found. This will only work on the continuous curve type. Quote Link to comment Share on other sites More sharing options...
lizp Posted May 18, 2009 Author Share Posted May 18, 2009 OK, I'll have too see how this can be depicted with continuous curves that can be handled to be of unusual shape. Wonder also if the sense of the torque driving vectors can be fixed in the upper left quadrant -- now they are placed in the opposite direction to what they should be? In terms of appearance it'll be good for the vectors to have small leads as well as broken lines (if the user wants) which connect the tips of the decomposed vectors with the tip of the vector they are derived from (that's not crucial but will make the drawing look nice). As for the curved line you drew, it looks great but one has to be able to reshape it in a desired way. It's interesting, the way you've drawn it leads to a torque on the order of 0.006Nm while the torques usually obtained are higher -- on the order of 0.009Nm, as seen from the graph. This seems to show how crucially important the form of that curve is. That curve is one of the main objects of optimization, I mentioned earlier, with the aim to obtain the higher net torque as possible at any position of the wheel. Quote Link to comment Share on other sites More sharing options...
Lee Mac Posted May 18, 2009 Share Posted May 18, 2009 I would model the curve as in Sean's example - as a spline. this can be modified very easily using control points. As for the upper left, I'll check it Quote Link to comment Share on other sites More sharing options...
Lee Mac Posted May 18, 2009 Share Posted May 18, 2009 Let me know if this solves the sign issue: Vector Analysis.lsp Quote Link to comment Share on other sites More sharing options...
lizp Posted May 18, 2009 Author Share Posted May 18, 2009 Great. We're almost there. The sense of the torque-generating vectors, which are the important ones, is correct and thus the sign of the individual torques seems to be correct in all quadrants. However, some of the component vectors along the normal (Green) in the lower left and the upper right quadrant, and the component parallel to the arm (Cyan) again in the lower left and the upper right quadrant -- are still directed in the reverse. These last vectors are not the main players but it's confusing visually if it's left that way. Quote Link to comment Share on other sites More sharing options...
lizp Posted May 18, 2009 Author Share Posted May 18, 2009 Oh, another minor thing, I forgot, when the labels are chosen to be placed manually each label is placed twice. Quote Link to comment Share on other sites More sharing options...
Lee Mac Posted May 18, 2009 Share Posted May 18, 2009 Hopefully this solves those problems Vector Analysis.lsp Quote Link to comment Share on other sites More sharing options...
lizp Posted May 18, 2009 Author Share Posted May 18, 2009 The label problem is solved but the normal component (Green) is reversed in the lower and upper left quadrants while the component along the arm (Cyan) is reversed in the lower and upper right quadrant. Quote Link to comment Share on other sites More sharing options...
Lee Mac Posted May 18, 2009 Share Posted May 18, 2009 Ha - I can't win with this thing Quote Link to comment Share on other sites More sharing options...
lizp Posted May 18, 2009 Author Share Posted May 18, 2009 As a general rule all the vectors, the base one and the derived have to be facing the same general direction, to put is loosely. Most of your vectors have the correct direction. Quote Link to comment Share on other sites More sharing options...
Lee Mac Posted May 18, 2009 Share Posted May 18, 2009 If this isn't right then I may just have to give up... Vector Analysis.lsp Quote Link to comment Share on other sites More sharing options...
Lee Mac Posted May 19, 2009 Share Posted May 19, 2009 OK, assuming the previous was correct, this includes highlighted components: Vector Analysis.lsp Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.