Ignouring the Relativity bad Pun: [img]/phpBB/images/smiles/icon_rolleyes.gif[/img]

Simple Answer:
Think of it like Pushing a Rock up a Mountain (Dang that Red Bull comercial), except that This Mountain is Special!

Although it Starts Off shallow, The Curve Steepens towards The End, with the Grade becoming 2/1 when you've gone 86.60% of The Way, 3/1 when you've reached 94.28%, [/b]7/1[/b] when you've hit 98.97%, and So On, and So Forth, with it Coming Closer to The Vertical, The Closer that you Get to The End, with it Becoming Infinitely Steep, not to be Confused with Vertical, by The End.

Big Nasty, Grab your Ears before your Brain Leaks out, Answer:

It's a Variation of The Pythagorean Theorem, known as The Lorentz-Fitzgerald Equation, after its Discoverers, it is Derived as Follows:

a^2=c^2-b^2

a=x=The Distance Traveled | b=v=The Velocity | c=The Speed of Light:

x^2=c^2-v^2

Divide both Sides by c^2:

x^2/c^2=1-v^2/c^2

Square Root both Sides:

x/c=SqRt(1-v^2/c^2)

This Version is used to Figure out Length Contraction in The Following Form | l'=The length Observed by the Moving Observer | l=The length Measured When at Rest with Respect to the Object:

l'=l*SqRt(1-v^2/c^2)

For Time, and Mass, Calculations, it Must be Inverted:

c/x=1/SqRt(1-v^2/c^2)

Often expressed in the Following Forms:

t'=The length Observed by the Moving Observer | t=The length Measured When at Rest with Respect to the Object:

t'=t/SqRt(1-v^2/c^2)

m'=The length Observed by the Moving Observer | m=The length Measured When at Rest with Respect to the Object:

m'=m/SqRt(1-v^2/c^2)

If we Set m Equal to 1, you Get The much Simpler:

m'=1/SqRt(1-v^2/c^2)

Just Remember though, you can Enter The Velocity as a Percentage, as it Saves you from Having to Type 299,792.458 km, All the Time!

Above All, HAVE FUN! [img]/phpBB/images/smiles/icon_biggrin.gif[/img]

_________________
If you Ignore YOUR Rights, they Will go away.

<font size=-1>[ This Message was edited by: ZaphodBeeblebrox on 2002-08-09 04:37 ]</font>