Date: 01/14/2001 at 02:37:08 From: Aaron Mangwiro Subject: Compound Angles I am installing a Ranch Rail vinyl fence with my Dad, and we have posts (5 x 5) with rails (2 x 3 1/2). The posts have 2.020 X 3.520 holes routed into them, and the rails fit into these holes perfectly. But we have one odd one. We have a situation where we have a post but the rail does not go into the post horizontally, but has a rise angle of 30 degrees and a side angle of 39 degrees. The hole to be routed needs to be a parallelogram instead of a square. I have tried to calculate the size of this parallelogram now for three weeks and asked every teacher, but no one could help me. Please help me - I am desperate. Thanks. Aaron
Date: 01/15/2001 at 09:51:02 From: Doctor Mitteldorf Subject: Re: Compound Angles Dear Aaron, Here's a formula that you can use: cosA*cosB = cosC. It applies to the three angles on planes that meet at a point, where two of the planes meet at right angles. Specifically: take a piece of cardboard, draw angle A, and cut it out. Take a second piece of cardboard and draw angle B. Now tape the pieces of cardboard together so the two vertices coincide, and one line of each angle is in common, and adjust the tape hinge so the cardboard planes are at right angles to one another. The third angle formed by the free edges of the cardboard is angle C, which obeys the formula above. (You can can convince yourself that this is true with some heavy-duty visualization, or by building a physical model. If you know vector algebra, the vector derivation is much easier.) Now picture the rail where it joins the plane of the post. Take as your vertex the point where the longest edge of the rail meets the post. Take angles A and B to be on the sides of the rail; they are (90-39) and (90-30) degrees. The third angle (C) is one angle of the parallelogram at the mouth of your hole in the post. By the formula above, cosC = cos(90-39)*cos(90-30), so C = 71.66 degrees, the acute angle in the parallelogram. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/
Date: 01/21/2001 at 13:12:22 From: Doctor Mitteldorf Subject: Re: Compound Angles Dear Aaron, Ever since you wrote this question to me, I've been going over it in my spare time and working on it. I gave you the wrong answer last week, and I think I have the right answer now. In the answer I gave you before, I assumed that the angles 30 and 39 degrees were line angles between the axis of the rail and the line orthogonal to the post. I don't think that's what you intended - I think the given angles are probably plane angles: first between the top face of the rail and the horizontal, second between the vertical face of the rail and the vertical plane perpendicular to the post. So I've re-done the analysis under that assumption. Last time, I started with the formula I call the hinge formula. As foundation, I want to clarify this and generalize it. Start with a hinge. On one arm of the hinge, draw an angle with the axis of the hinge as one side. Call this angle alpha. On the other arm of the hinge, draw angle beta, again with the hinge axis as one side. You now have two angles in different planes, sharing a common ray. Open the hinge to a right angle. Measure the angle gamma between the outer rays of angles alpha and beta. Then the three angles are related by cos(gamma) = cos(alpha)*cos(beta). This is the hinge formula for right angle opening. In general: Suppose the hinge is open to an arbitrary angle zeta. Then the formula becomes cos(gamma) = cos(alpha)*cos(beta) + sin(alpha)*sin(beta)*cos(zeta) This is the general hinge formula for arbitrary opening angle zeta. Now imagine three planes that come together at a point. They form three plane angles and three line angles, and you can apply the hinge formula in three different ways. Just as you can write cos(gamma) in terms of alpha and beta, you can also write cos(alpha) in terms of beta and gamma, and you can write cos(beta) in terms of gamma and alpha. In fact, as the hinge formula is set up, it gives you one line angle in terms of the other two line angles and the opposite plane angle; but any three of the angles are sufficient to solve the system. In your case, I believe you have started with a rail aligned along a horizontal axis, with its top face horizontal and its side faces vertical and perpendicular to the rail. You've turned the rail horizontally through an angle of 39 degrees, and then tilted it down through an angle of 30 degrees. So consider the point where the longest edge of the rail meets the rail. There are three faces coming together at that point: two of them are faces of the rail, and the third is the vertical face of the post. The two rail faces, of course, meet each other at a right angle. The horizontal face of the rail meets the vertical face of the post at an angle 90-30 = 60 degrees, and the vertical face of the rail meets the vertical face of the post at an angle 90-39 = 51 degrees. So in this case, you have the three plane angles, and you'd like to find the line angles - in particular, the angle between the lines in the parallelogram, which you need to rout in the vertical face of the post. So we go back to these three hinge formulas, specialized where one face angle is a right angle. We can solve the three equations for the three line angles. Call the three line angles alpha, beta, and gamma; call the two plane angles theta and phi - the third is 90 degrees. The three hinge formulas are: 1) cos(alpha)=cos(gamma)*cos(beta) + sin(gamma)*sin(beta)*cos(theta) 2) cos(beta)=cos(gamma)*cos(alpha) + sin(gamma)*sin(alpha)*cos(phi) 3) cos(gamma)=cos(alpha)*cos(beta) You can solve this system for alpha, beta, and gamma, given theta and phi. Here are some hints how to do this: 1) Write x for cos(alpha), y for cos(beta) and z for cos(gamma). Then sin(alpha)=sqrt(1-x^2), etc. 2) Substitute equation (3) for z into equations (1) and (2) to give two equations in x and y. Solve either one, and substitute into the other. Some fortunate simplifications are available along the way, and the result is surprisingly simple: cos(alpha) = cos(theta)/sin(phi) cos(beta) = cos(phi)/sin(theta) gamma is the angle in the parallelogram that you're looking for. cos(alpha) = cos(51)/sin(60) ==> alpha = 43.4 deg cos(beta) = cos(60)/sin(51) ==> beta = 50.0 deg cos(gamma) = cos(alpha)*cos(beta)=0.468 ==> gamma = 62.1 deg - Doctor Mitteldorf, the Math Forum
Date: 01/24/2001 at 13:04:10 From: Dennis Mangwiro Subject: Re: Compound Angles Thank you very much - that was amazing. I have one more question. Will the width of the parallelogram be the same, taking into consideration the wall thickness of 0.150 inch, as the width of the the widened rectangle due to the side angle only? Similarly, will the length of the parallelogram side be the same as the length of the side of the rectangle, enlarged, due to the rise angle only? Thank you. Aaron
Date: 01/25/2001 at 06:52:12 From: Doctor Mitteldorf Subject: Re: Compound Angles Dear Aaron, Last time, I described a method for calculating the three line angles where three planes meet. The three planes we're looking at here are the top and side of the rail and the vertical face of the post. When you look at the point where the longest edge of the rail meets the post, one of the line angles we calculated gives you the acute angle in the parallelogram traced on the face of the post. The other two line angles, which we calculated as alpha and beta, appear on the top and the vertical side of the rail. In fact, you can draw a picture for yourself and see that the length and width of the rectangular cross-section of the rail are elongated by factors of 1/cos(alpha) and 1/cos(beta). Does alpha go with the length and beta with the width or vice versa? I'll leave that to you to figure out. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/
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