beero1000
Veteran Member
I think you might be off by a factor of -1. I get cos(b) = (cos(a) - 1)/(cos(a) + 1) for a square pyramid.
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I think you might be off by a factor of -1. I get cos(b) = (cos(a) - 1)/(cos(a) + 1) for a square pyramid.
Jokodo
Veteran Member
I am impressed by your thinking of that as a dumb question. I can think of dumber ones. You are talking about an intermediate thread at least, surely.
But I'll play. My dumb question is....how is the answer not always 90o regardless of the top angle?
Think of the extreme cases: For an flat "pyramid", where the angles at the top are 90o (for a four sided pyramid), and thus the pyramid is just a plane parallel and coincident with the base, the angle is 180o. Only for an infinitely steep pyramid (with angles 0o at the top) will it be 90o).
Kharakov
Quantum Hot Dog
Yeah, but there are many unique infinities. (1-1)^-1 = 1+1+1+1+1.. is different compared to (1-1)^-2 = 1+2+3+4...
beero1000
Veteran Member
Figured I would share the general calculations for dihedral angles.
Suppose we have two planes (faces) F and G that intersect at (share the edge) E. Choose 2 points A and B on E, a point C on F, and a point D on G. To make it easier on ourselves, we'll assume that B, C, and D are all at unit distance from A, which will be the origin.
Then, the angle X between the faces is equal to the angle between normals N and M of F and G, where the normals are given by the appropriate cross products:
N = AB × AC
M = AB × AD
We can get the angle between N and M with a dot product, using the standard cosine formula for the dot product N · M = |N| |M| cos X and the standard sine formula for the cross product:
N · M = |N| |M| cos X = sin(BAC) sin(BAD) cos(X)
Additionally, we can get the same dot product with the Cauchy identity:
N · M = (AB × AC) · (AB × AD) = (AB · AB) (AC · AD) - (AB · AD) (AC · AB) = cos(CAD) - cos(BAD) cos(CAB)
Putting all of that together gives:
cos(X) = (cos(CAD) - cos(BAD) cos(CAB))/(sin(BAC) sin(BAD))
In our instance of a pyramid with all angles at vertex A equal to Y, we get:
cos(X) = (cos(Y) - cos2(Y))/sin2(Y) = tan(Y/2)cot(Y) if the pyramid has a triangular base
and
cos(X) = ((2cos(Y) - 1) - cos2(Y))/sin2(Y) = - tan2(Y/2) = (cos(Y) - 1)/(cos(Y) + 1) if the pyramid has a square base
Suppose we have two planes (faces) F and G that intersect at (share the edge) E. Choose 2 points A and B on E, a point C on F, and a point D on G. To make it easier on ourselves, we'll assume that B, C, and D are all at unit distance from A, which will be the origin.
Then, the angle X between the faces is equal to the angle between normals N and M of F and G, where the normals are given by the appropriate cross products:
N = AB × AC
M = AB × AD
We can get the angle between N and M with a dot product, using the standard cosine formula for the dot product N · M = |N| |M| cos X and the standard sine formula for the cross product:
N · M = |N| |M| cos X = sin(BAC) sin(BAD) cos(X)
Additionally, we can get the same dot product with the Cauchy identity:
N · M = (AB × AC) · (AB × AD) = (AB · AB) (AC · AD) - (AB · AD) (AC · AB) = cos(CAD) - cos(BAD) cos(CAB)
Putting all of that together gives:
cos(X) = (cos(CAD) - cos(BAD) cos(CAB))/(sin(BAC) sin(BAD))
In our instance of a pyramid with all angles at vertex A equal to Y, we get:
cos(X) = (cos(Y) - cos2(Y))/sin2(Y) = tan(Y/2)cot(Y) if the pyramid has a triangular base
and
cos(X) = ((2cos(Y) - 1) - cos2(Y))/sin2(Y) = - tan2(Y/2) = (cos(Y) - 1)/(cos(Y) + 1) if the pyramid has a square base
Kharakov
Quantum Hot Dog
Wolfram alpha heads, can someone please help me with the following?
\(\sum_{n=1}^{k}1 - \frac{n}{2} + \frac {(n-1)n(n+1)}{4!3!}- \frac{(n-2)(n-1)n(n+1)(n+2)}{6!5!} + \frac{(n-3)(n-2)(n-1)n(n+1)(n+2)(n+3)}{8!7!}- \, ...\)
by help I mean.. how do I input the double sum?
nm, I can simplify it.. times up on editing though
\(\sum_{n=1}^{k}1 - \frac{n}{2} + \frac {(n-1)n(n+1)}{4!3!}- \frac{(n-2)(n-1)n(n+1)(n+2)}{6!5!} + \frac{(n-3)(n-2)(n-1)n(n+1)(n+2)(n+3)}{8!7!}- \, ...\)
by help I mean.. how do I input the double sum?
nm, I can simplify it.. times up on editing though
Last edited:
Kharakov
Quantum Hot Dog
\(\sum_{n=1}^{k}1 - \frac{n}{2} + \frac {(n-1)n(n+1)}{4!3!}- \frac{(n-2)(n-1)n(n+1)(n+2)}{6!5!} + \frac{(n-3)(n-2)(n-1)n(n+1)(n+2)(n+3)}{8!7!}- \, ... \)
=
\(1 - \frac{k(k+1)}{2!2!} + \frac {(k-1)k(k+1)(k+2)}{4!4!}- \frac{(k-2)(k-1)k(k+1)(k+2)(k+3)}{6!6!} + \frac{(k-3)(k-2)(k-1)k(k+1)(k+2)(k+3)(k+4)}{8!8!}- \, ...= \)
=
\(1 - \frac{k(k+1)}{2!^2} + \frac {(k-1)k(k+1)(k+2)}{4!^2}- \frac{(k-2)(k-1)k(k+1)(k+2)(k+3)}{6!^2} + \frac{(k-3)(k-2)(k-1)k(k+1)(k+2)(k+3)(k+4)}{8!^2}- \, ...= \)
=
\(1 - \frac{k(k+1)}{2!2!} + \frac {(k-1)k(k+1)(k+2)}{4!4!}- \frac{(k-2)(k-1)k(k+1)(k+2)(k+3)}{6!6!} + \frac{(k-3)(k-2)(k-1)k(k+1)(k+2)(k+3)(k+4)}{8!8!}- \, ...= \)
=
\(1 - \frac{k(k+1)}{2!^2} + \frac {(k-1)k(k+1)(k+2)}{4!^2}- \frac{(k-2)(k-1)k(k+1)(k+2)(k+3)}{6!^2} + \frac{(k-3)(k-2)(k-1)k(k+1)(k+2)(k+3)(k+4)}{8!^2}- \, ...= \)
Jokodo
Veteran Member
That's fairly easy for me, because I have Mathematica. It does algebra, but I can do analytic geometry with it, even if not Euclid-style constructions.
h = (height) / (half of base length)
cos(angle between edges at the top of the pyramid: a) = h2 / (1 + h2)
cos(angle between sides: b) = 1 / (1 + 2*h2)
h = sqrt(cos(a)/(1 - cos(a))
cos(b) = (1 - cos(a))/(1 + cos(a))
Orbital elements has the solution to that problem.
cos(asn)*cos(pla) - sin(asn)*sin(pla)*cos(inc)
sin(asn)*cos(pla) + cos(asn)*sin(pla)*cos(inc)
sin(pla)*sin(inc)
asn = angle from {1,0,0} in the equator to the ascending node
inc = inclination of orbit plane to equator
pla = angle from the ascending node to the object position
I think what you gave me here is the direction from the origin to the object? I got that already, or a good approximation for it. What I was actually looking for, and my wording may have been imprecise, is the direction of movement.
lpetrich
Contributor
Take the derivative of the position vector as a function of time.
Taking it for pla gives the perpendicular part of the direction of motion:
- cos(asn)*sin(pla) - sin(asn)*cos(pla)*cos(inc)
- sin(asn)*sin(pla) + cos(asn)*cos(pla)*cos(inc)
sin(pla)*cos(inc)
For position vector x = r*n for radius r and direction vector n,
dx/dt = (dr/dt)*n + (dn/da)*(r*(da/dt)) = vr*n + vt*(dn/da) for radial and transverse velocities vr and vt.
lpetrich
Contributor
\(\sum_{n=1}^{k} \sum_{m=0}^\infty (-1)^m \cdot \frac{(n-m)(n-m+1) \cdots (n+m-1)(n+m)}{(2m)!(2m-1)!} \)
or
\(\sum_{n=1}^{k} \sum_{m=0}^\infty (-1)^m \cdot \frac{(n+m)!}{(n-m-1)!(2m)!(2m-1)!} \)
Jokodo
Veteran Member
Unless I'm misunderstanding something, this can't be correct. Just looking at some of the extreme cases, e.g. when pla is 90°, z will always be zero no matter the inclination - when the object reaches the highest (or lowest) point in it's orbit, it'll temporarily show no vertical motion, and even if the inclination is 90°, there'll be no vertical movement in the moment the object crosses over the pole.
I'm afraid I'll have to look at this more closely when the kid is in bed, this is too high to grasp at a glance.
ruby sparks
Contributor
I am impressed by your thinking of that as a dumb question. I can think of dumber ones. You are talking about an intermediate thread at least, surely.
But I'll play. My dumb question is....how is the answer not always 90o regardless of the top angle?
Think of the extreme cases: For an flat "pyramid", where the angles at the top are 90o (for a four sided pyramid), and thus the pyramid is just a plane parallel and coincident with the base, the angle is 180o. Only for an infinitely steep pyramid (with angles 0o at the top) will it be 90o).
Thx. So the angle between the faces at those two extremes is angle at top plus 90 degrees....and I'm wondering why the angle between the faces should not change linearly in proportion to a linear change in the angle at the top and therefore be angle at top plus 90 degrees for all top angles in between (0 and 90).
Which for angle at top of 45 would give angle between faces of 135. Etc.
Also...someone mentioned an angle at top of 120. But could you make a square-based pyramid with 4 triangles like that?
lpetrich
Contributor
I made a typo. Here is the correct form:
- cos(asn)*sin(pla) - sin(asn)*cos(pla)*cos(inc)
- sin(asn)*sin(pla) + cos(asn)*cos(pla)*cos(inc)
cos(pla)*sin(inc)
Notice that this direction is 90d ahead of the position direction in the plane of the orbit.
beero1000
Veteran Member
It can't be linear because a small change when it almost flat will have a much greater effect than a small change when the angles are very small. You can see this effect in folding tripods.
The sum of the angles around a convex polyhedral vertex must be less than 360 degrees. If it's a pyramid with a triangular base that's a limit of three 120 degree angles, and if it's a pyramid with a square base that's a limit of four 90 degree angles
Jokodo
Veteran Member
Thanks!
I'm aware that it has to be 90d ahead in the plane of the orbit. I just never got the hang of translating between coordinate systems.
bilby
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Jokodo
Veteran Member
I did call it a dumb math question...
beero1000
Veteran Member
Why not t = 2pi/log(N)?
lpetrich
Contributor
"Natural number" is ambiguous. I've seen two definitions; "positive integer" and "nonnegative integer". The second is the first with zero added.
Since these two sets of numbers are subsets of the real numbers, I will work with real numbers because the proof is somewhat easier.
Negative real numbers -x: log(-x) = log(x) + pi*i
Zero: log(0) = - infinity
So we are left with positive real numbers. The problem becomes: for some real y such that y = log(x), is x unique?
Solution: Let us suppose that there are two positive real numbers, x1 and x2, such that log(x1) = log(x2). By Rolle's theorem, the derivative of the logarithm function must be 0 for at least one x such that x1 <= x <= x2. That derivative is 1/x, and it is nonzero for all positive x. Thus, there can no such pair of numbers, and each real y gives a unique positive real x.
Since these two sets of numbers are subsets of the real numbers, I will work with real numbers because the proof is somewhat easier.
Negative real numbers -x: log(-x) = log(x) + pi*i
Zero: log(0) = - infinity
So we are left with positive real numbers. The problem becomes: for some real y such that y = log(x), is x unique?
Solution: Let us suppose that there are two positive real numbers, x1 and x2, such that log(x1) = log(x2). By Rolle's theorem, the derivative of the logarithm function must be 0 for at least one x such that x1 <= x <= x2. That derivative is 1/x, and it is nonzero for all positive x. Thus, there can no such pair of numbers, and each real y gives a unique positive real x.
Kharakov
Quantum Hot Dog
So every number will have an increasing cycle length before it is again 0 mod 2pi. 2 to 4 to 8 to 16... 3,9,27,81... etc..