ID:181731
 Oct 19 2009, 11:35 am
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Anyone have the math for the handy? I can't seem to find it.
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Oct 19 2009, 11:40 am
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Airjoe
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What are your knowns?
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What values do you know? Just the bases and the height, or the lengths of the sides as well? Angles?
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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.
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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? | |
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 | |
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 [Edit] 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) | |
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: ... Nevermind. Too darn complicated for my inferior math skills. Isn't there a library for this somewhere? | |
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. | |
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: 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 | |
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. | |
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. | |