ID:181731   Oct 19 2009, 10:35 am Anyone have the math for the handy? I can't seem to find it.
 <-> Oct 19 2009, 10:40 am What are your knowns?
 <-> Oct 19 2009, 10:40 am In response to Airjoe The whats?
 <-> Oct 19 2009, 10:42 am In response to Foomer What values do you know? Just the bases and the height, or the lengths of the sides as well? Angles?
 <-> Oct 19 2009, 10:43 am In response to Airjoe Just length and width is all I need. The center of a 2D lopsided square. I figure once you know how to do it in 2D you can easily do it in 3D anyway if needed.
 <-> Oct 19 2009, 10:44 am In response to Foomer I'm saying, what values do you know beforehand? Click For instance, in that image, do you know lengths AD, DC, CB, AB? What about h?
 <-> Oct 19 2009, 10:51 am In response to Airjoe You've got 4 points, say: 55x, 14y 55x, -14y -55x, 31y -55x, -31y
 <-> Oct 19 2009, 11:01 am (Edited on Oct 19 2009, 11:08 am) In response to Foomer Probably your easiest approach is to break it into triangles, which also simplifies the problem of what to do if two sides aren't parallel. Find the centroid and area of one triangle, then the centroid and area of the other, and get a weighted average. To get the centroid of a triangle you find the intersection of its medians, so you'd take a line from halfway along one vector and draw it to the opposite corner, and then do that with a different vector, find the intersection, and there you are. I can't remember the area formula in terms of vectors for the life of me, but I'd bet it's related to the cross-product. (Actually check that; the area is the cross-product, the absolute value of it at least. Doh.) Lummox JR
 <-> Oct 19 2009, 11:15 am (Edited on Oct 19 2009, 11:40 am) In response to Foomer http://www.efunda.com/math/areas/Trapazoid.cfm In that diagram, A is 28 (14-(-14)), B is 62 (31-(-31)), C is 17 (31-14), h is 110 (55-(-55)). With these values A, B, C, h, we can find the center as: Cx = (2ac+a^2+cb+ab+b^2)/(3(a+b)) Cy = (h(2a+b))/(3(a+b)) More difficult than I thought it was. So, we can compute: Cx = (2*28*17 + 28^2 + 17*62 + 28*62 + 62^2)/(3*(28+62)) = 31 Cy = (110(2*28+62))/(3*(28+62)) = 48.0740741 These values Cx, Cy, are from point -55, 31  Yeah, Cy is right, because it's going to be a little more left leaning than "centered". So, in the end, the Center is at location (6.9259259,0)
 <-> Oct 19 2009, 12:16 pm In response to Airjoe Airjoe wrote: A is 28 (14-(-14)), B is 62 (31-(-31)), C is 17 (31-14), h is 110 (55-(-55)). With these values A, B, C, h, we can find the center as: Cx = (2ac+a^2+cb+ab+b^2)/(3(a+b)) Cy = (h(2a+b))/(3(a+b)) Cx = (2*28*17 + 28^2 + 17*62 + 28*62 + 62^2)/(3*(28+62)) = 31 Cy = (110(2*28+62))/(3*(28+62)) = 48.0740741 ... Nevermind. Too darn complicated for my inferior math skills. Isn't there a library for this somewhere?
 <-> Oct 20 2009, 11:56 am In response to Airjoe It also appears that I need to calculate the Center of Gravity, not just the center. (Apparently there's a difference.) I've found a site that explains how to do this: http://www.engineeringtoolbox.com/center-gravity-d_1310.html I just don't know how to translate what its saying to do into math.
 <-> Oct 20 2009, 12:09 pm In response to Foomer Foomer wrote: It also appears that I need to calculate the Center of Gravity, not just the center. (Apparently there's a difference.) I've found a site that explains how to do this: http://www.engineeringtoolbox.com/center-gravity-d_1310.html I just don't know how to translate what its saying to do into math. The formulas mentioned here will suffice, unless your trapezoids are not of uniform density (e.g., varying thickness of the material from one end to the other). When density is uniform, the geometric center is the same as the center of mass. Lummox JR
 <-> Oct 20 2009, 12:15 pm In response to Lummox JR I'm not sure what the formulas mentioned here result in, but from the look of the picture on that site it'll end up something like this: Now that may be mathematically the center of the shape, but it certainly is not the center of gravity, since if that were a piece of wood and you attempted to balance it at the intersection, the south side would be notably heavier than the north side.
 <-> Oct 20 2009, 3:35 pm In response to Foomer For points: -55, 31 -55, -31 55, 14 55, -14 The center of gravity is located at: -7,0 It's not exactly -7, but just about. The horizontal line you drew is not infact the "mathematical center", just half of the height.
 <-> Oct 21 2009, 1:31 pm In response to Airjoe Sorry, what you're doing to get these numbers makes absolutely no sense to me.
 <-> Oct 21 2009, 1:36 pm In response to Foomer Do you have some form of IM in which I can walk you through it? Are you opposed to joining Chatters?
 <-> Oct 23 2009, 11:15 am In response to Lummox JR Do you know how to determine the coordinates of an intersection? If you have four points, and you draw two lines between them making an X, how do you determine where the intersection is?