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

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.
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Approach your problem from the right end and begin with the answers. Then one day, perhaps you will find the final question..
R. van Gulik

Posts: 23951
Joined: September 20th, 2002, 8:30 pm

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

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.

Cuchulainn
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Posts: 60753
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
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### Re: Going off on a tangent (kind of) : find y

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 $\theta$?
http://www.datasimfinancial.com
http://www.datasim.nl

Approach your problem from the right end and begin with the answers. Then one day, perhaps you will find the final question..
R. van Gulik

ppauper
Posts: 70239
Joined: November 15th, 2001, 1:29 pm

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

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

Cuchulainn
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Posts: 60753
Joined: July 16th, 2004, 7:38 am
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### Re: Going off on a tangent (kind of) : find y

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?
http://www.datasimfinancial.com
http://www.datasim.nl

Approach your problem from the right end and begin with the answers. Then one day, perhaps you will find the final question..
R. van Gulik

Cuchulainn
Topic Author
Posts: 60753
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
Contact:

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

1. Cool way, isn't it? to find power series for tan (t).
http://www.datasimfinancial.com
http://www.datasim.nl

Approach your problem from the right end and begin with the answers. Then one day, perhaps you will find the final question..
R. van Gulik

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