Precession of Mercury's longitude of perihelion - (so called "anomaly")

On Sun, 27 May 2007 08:55:16 GMT, "Dirk Van de moortel"
wrote:


wrote in message ups.com...

Dirk Van de moortel wrote:
wrote in message oups.com...
Assume measurement accuracy to "within 43 arc seconds" per century.

Assume 1 Earth century = 415 Mercury orbits

~~~~~~

415 orbits * 360 degrees = 149400 degrees per century
149400 degrees * 60 arc minutes = 8964000 arc minutes per century
8964000 arc minutes * 60 arc seconds = 537840000 arc seconds per
century.


43
-------------------------------- = 0.0000000799494273389855719 =
7.9949E-8
537840000

Are we all on the same page so far?

Sure.


If so, what the hell am I missing when 7.9949E-8 parts out of 100 is
not equal to 7.9949E-6 percent?
N parts out 100 is N %. N = N.
It is, but that is not what he is saying.
7.9949E-8 times 100 is not 7.9949E-6 percent
It is
7.9949E-6
and it is
7.9949E-4 percent


Am I suffering from a brain embolism here or something?? 'Cause it's
either that or I'm completely missing the point of
http://users.telenet.be/vdmoortel/di...rcentages.html.

For the love of god, help!

The point is that the engineer doesn't understand what percentages are.
Do you understand the meaning of the percent-sign?

Dirk Vdm

Hah...obviously not.

% is an abreviation of 1/100


Assume a 15 question quiz. You answer 11 correct out of 15. Your
percentage score is:

11/15 = .73/1 = x/100.

Yes.
11/15 = 11/15 * 1
= 11/15 * 100 %
= 73.1 %
and that should be all there is to it.

Dirk Vdm