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[Descan]

I'm doin' some trig review, and maybe someone can help me out here.

It's about cosine law, and I got this:

Sides are, a = 4, b = 7, c = 8, find angle C in degrees.

The answer I got is 91.02, because 64 - 49 - 16 = -1.

The answer is actually 88.98, which I get if I use 1 instead of -1.

I'm havin' trouble quite grokking why? Is it because either one works, it just depends on whether it's an acute or obtuse angle at C? If so, is there like a handy rule of "If they ask this kind of thing, use the smaller angle" or do I have to just guess? :V

[Jim Groovester]

I get the right answer doing the law of cosines. You must be making an arithmetic error somewhere.

[Descan]

Spoiler (click to show/hide)

c² = a² + b² - 2abcos(C)
8² = 4² + 7² - 2(4)(7)cos(C)
64 = 16 + 49 - 56cos(C)
64 - 16 = 49 - 56cos(C)
48 - 49 = -56cos(C)
-1 = -56cos(C)
1/56 = cos(C)
cos^-11/56 = (C)
88.98^deg = (C)

Yup. Figured it out. Dropped a negative in the bolded area.

[DreamThorn]

I have to get my math skills back up to scratch, so I'm trying to (re)derive Newton's solution to the two body problem.

I managed to get:

x'' = -GMx / ||x||³

with '' being the second time derivative, x being the vector distance between the two masses as a function of time, and M being the sum of the masses.

Which is what I expected to get, but now I have to find x.  As far as I know, this is done by guessing forms for x and finding one that fits, and the important clue is that x'' is dependent on -x.

So I tried x(t) = K.exp(ivt+b), with K, v and b are unknown constant vectors and i² = -1, because I think this will have enough freedom for circular, elliptical, parabolic and hyperbolic orbits.

x'' = -v².K.exp(ivt+b) = -v².x        (with v² being v with each element squared; not v dot v.  Similarly, the full stops are per element multiplication, not vector multiplication)
-v².x = -GMx / ||x||³

So either x = 0, or
-v² = -GM / ||x||³
||x||³ = GM/ v²        (So the distance between the masses remains constant, and v must have the same value for all elements)

I think that bringing complex numbers in might have been a mistake, I'll try ordinary trig functions next.

P.S. I mostly wrote this thing so that I could analyze my thought processes, but am posting it in case someone else enjoys reading it.

[Gigaz]

This result is not very surprising imo. For an elliptical orbit around a heavy object the correct Newtonian solution is the one where the heavy object is in one of the focal points. Your assumption only allows for ellipsoids with a center of mass at x=0.

[Eagle_eye]

Does anyone know of a good source for learning tensor analysis that I could acquire without paying money?

[DreamThorn]

It seems my mistake was actually that I ignored that x'' is also related to ||x||³, not just x.

I have managed to show now that at least a circular orbit works.  The equations for the other options will have to take into account that the objects speed up as they get closer and slow down as they separate, so these will be more difficult.

@Gigaz: Here x is the difference in position of the two bodies;  a part of the calculation that I did not show, but did first, proved that this is equivalent to one mass being stationary, even though both bodies actually orbit the center of mass.

[frostshotgg]

All right, applied mathematics help. I'm making a Evrard Ventilator, I'm having issues figuring out the equations for the tooth and cut-in. Based on the only documentation I can find (p439 in the ebook), I know each half of the tooth is a cardioid, and I know the cut-in is an epitrochoid, but I have no idea what sections of the graph it should be, among other problems. Any ideas?

[DreamThorn]

I think the key to figuring that out is understanding why they are cardioids and epitrochoids, specifically.  That should lead to understanding the necessary properties of the curves and then their construction.

I'll keep thinking on this.

EDIT:

I think it goes like this:

1. Pick radii for the two circles. Call these R_in and R_out.  One must be a whole multiple of the other.
2. Pick the depth of the groove.  Call this r.
3. The epitrochoid is such that a circle with radius R_out is rolling on a circle with radius R_in and the traced point is distance (R_out + r) from the center of the rolling circle.
4a. The cardioids are both centered on the circle with radius R_out.
4b. The cardioids are such that their respective sharp points are at the points of the R_out circle that will touch the intersection of the epitrochoid and the R_in circle...
4c. and that they intersect each other at a distance of (R_out + r) from the center of the circle with radius R_out.

[MagmaMcFry]

All right, applied mathematics help. I'm making a Evrard Ventilator, I'm having issues figuring out the equations for the tooth and cut-in. Based on the only documentation I can find (p439 in the ebook), I know each half of the tooth is a cardioid, and I know the cut-in is an epitrochoid, but I have no idea what sections of the graph it should be, among other problems. Any ideas?

That's easy to explain. Just imagine one part being held still while the other part rolls around it. The epitrochoid is cut by the tip of the teeth, and the cardioid is cut by the edges of the groove.

[DreamThorn]

That's what I said.

[frostshotgg]

Yes, I can process what's going on. I can't process what the equations are to describe it.

[ZetaX]

If we assume that 2^{dim F(a^i)] is simply the number of fixed points of a^i, then the result as written follows simply by Möbius inversion.
Thus if we take F(a^i) to be the eigenspace of a^i corresponding to the eigenvalue 1, then everything fits.

Also, the proof and statement of that theorem is unnexessarily weird and long.

[ZetaX]

Calculate the Eigenspace with any method you want; you essentially need to solve the equation a^i v = v, which is homogeneous and linear.

If you want to use that formula to calculate the number of p-loops, then you would factorize p first, then calculate a^d for all divisors d of p (use fast exponentiation; you can also do it iteratively), then find the eigenspace dimensions (Gauß' algorithm or whatever else that works), then apply the formula.

[ZetaX]

You are working in IZ/2 aka IF_2 aka the field with two elements (can't better format that here, I used I to denote a second stroke both times): 2 is the same as 0 there, and thus 1 is the same as -1, too.