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### Re: Going off on a tangent (kind of) : find y

Posted: May 20th, 2017, 8:13 pm
Seems Kepler was a very naughty boy..

http://www.nytimes.com/1990/01/23/scien ... gewanted=1

Done in 1609, Kepler's fakery is one of the earliest known examples of the use of false data by a giant of modern science.

### Re: Going off on a tangent (kind of) : find y

Posted: May 21st, 2017, 1:01 am
Kepler is not good enough for GPS -- they need Einstein to get the accuracy.
Interesting. Do you have a link?
Actually, up at Myvatn was a retired American prof talking about his day with Atlas (D?) in the 50s/60s and the very issue of time synchronisation. They let the satellite broadcast the 'time'. There is no day or night up there.
Here's one: http://www.astronomy.ohio-state.edu/~po ... 5/gps.html

The showstopper if relativity is ignored in GPS:
If these effects were not properly taken into account, a navigational fix based on the GPS constellation would be false after only 2 minutes, and errors in global positions would continue to accumulate at a rate of about 10 kilometers each day! The whole system would be utterly worthless for navigation in a very short time.

### Re: Going off on a tangent (kind of) : find y

Posted: May 21st, 2017, 12:19 pm
Having found [$]y[$] you now need to solve for the true anomaly [$]\theta[$], How?

[$]\tan^2(\theta/2) = a \tan^2(y/2)[$] where [$] a = (1+\varepsilon)/(1 - \varepsilon)[$].

What's [$]\theta[$]?

### Re: Going off on a tangent (kind of) : find y

Posted: May 21st, 2017, 12:34 pm
Having found [$]y[$] you now need to solve for the true anomaly [$]\theta[$], How?

[$]\tan^2(\theta/2) = a \tan^2(y/2)[$] where [$] a = (1+\varepsilon)/(1 - \varepsilon)[$].

What's [$]\theta[$]?
you keep changing this, there have been at least 3 different expressions since I started to answer it

[$]\tan^2(\theta/2) = a \tan^2(y/2)[$]  so [$]\theta=\pm2\arctan\left[\sqrt{a \tan^2(y/2)}\right][$], with the addition of some multiple of [$]2\pi[$] if necessary for physical reasons

### Re: Going off on a tangent (kind of) : find y

Posted: May 21st, 2017, 1:19 pm
Sorry, I wanted to write the unknown on the left. Looks like a whole bunch of solutions. I suppose only one of them is physically correct?

### Re: Going off on a tangent (kind of) : find y

Posted: October 22nd, 2017, 9:27 am
1. Cool way, isn't it? to find power series for tan (t).