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

by Traden4Alpha

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.

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

by Cuchulainn

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

by ppauper

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

by Cuchulainn

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

by Cuchulainn

Cool way, isn't it? to find power series for tan (t).