Cross-Cornering a Shape to Make it SquareDate: 12/19/2002 at 22:52:31 From: Tim Barker Subject: Cross cornering a shape to make it square At my work we lay out buildings. Lets say we lay out a building that is 30' x 40'. When you cross corner it to see if it is "square" there is a 6" difference. What is the equation to find how far to move one side to make the shape "square"? Date: 12/20/2002 at 10:41:47 From: Doctor Rick Subject: Re: Cross cornering a shape to make it square Hi, Tim. This is a good question! It will take a lot of math to derive the answer; feel free to skip down to the end to see the formula I come up with. In geometric terms, if you've measured so all the sides are the correct length (opposite sides are the same length), then you have a parallelogram, and you want to know how far the angles are from right angles, based on the difference in the diagonals. We can use the formulas for parallelograms in the Dr. Math FAQ: http://mathforum.org/dr.math/faq/formulas/faq.quad.html There we find formulas for the diagonals p and q in terms of angles A and B. (Diagonal p joins the vertices with angle B, and diagonal q joins the vertices with angle A; a and b are the two side lengths.) p = sqrt[a^2+b^2-2ab cos(A)] q = sqrt[a^2+b^2-2ab cos(B)] We know that angles A and B are supplementary, that is, their sum is 180 degrees. Thus B = 180-A, and cos(B) = cos(180-A) = -cos(A). Now we can express the difference in the diagonals in terms of A: q-p = sqrt[a^2+b^2+2ab cos(A)] - sqrt[a^2+b^2-2ab cos(A)] It's quite a chore to solve this equation for the unknown angle A. But I'll assume that the difference is small, like 6 inches out of 50 feet. Then the angle is close to 90 degrees, and the cosine of the angle is approximately equal to the difference between A and 90 degrees, converted to radians. (180 degrees equal pi radians, that is, 3.14159... radians.) Calling this angle difference x, the equation becomes q-p = sqrt(a^2+b^2+2abx) - sqrt(a^2+b^2-2abx) It's still tough, but I can make another approximation. Since x is small, I can factor out the term (a^2+b^2): q-p = sqrt(a^2+b^2)(sqrt(1+2abx/(a^2+b^2))-sqrt(1-2abx/(a^2+b^2) and approximate sqrt(1+e) as (1+e/2) where e = +-2abx/(a^2+b^2) is a small quantity: q-p = sqrt(a^2+b^2)(1+abx/(a^2+b^2)-1+abx/(a^2+b^2)) = sqrt(a^2+b^2)*2abx/(a^2+b^2) = 2abx/sqrt(a^2+b^2) Now we can solve for x: x = (q-p)sqrt(a^b+b^2)/(2ab) There's one more step. You probably aren't interested in the angle, but in how far you need to move one end of a side. Let's say you want to move side b. The distance you need to move it is b times the angle in radians, which is x. Thus distance = bx = (q-p)*sqrt(a^2+b^2)/(2a) A final simplification uses the fact that sqrt(a^2+b^2) is pretty close to the average of the diagonals, (p+q)/2. distance = (q-p)((p+q)/2)/(2a) YOU MAY SKIP TO HERE: In words: Multiply the difference in the diagonals by the average of the diagonals. Then divide by the twice the length of the side you're not moving. This is the distance to move. If the diagonal from the stationary end of this side is the longer diagonal, you'll want to move the side out (increasing angle A); if this diagonal is the shorter diagonal, move the side in. In your example, if you want to move side b: side a = 40 feet side b = 30 feet difference of diagonals, q-p = 6 inches = 0.5 foot average of diagonals, (p+q)/2 = 50 feet distance to move = 0.5 * 50 / 80 = 0.3125 feet = 3.75 inches I'll just check this result. If I was 0.3125 feet off, the angle was off by (0.3125/30)*180/pi = 0.5968 degrees; the angles in the parallelogram were A = 89.4032 and B = 90.5968 degrees. Using the formulas for the diagonals, I calculate p = 49.749 feet and q = 50.249 feet. The difference in the diagonals is thus 0.500 foot. The formula worked. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/