Gravity made easy

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Postby Natalya » Thu Dec 04, 2008 2:18 pm

Taking trianglism to a whole new level...
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Postby pesgores » Thu Dec 04, 2008 2:23 pm

Rody wrote:Image

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Postby Rody » Fri Dec 05, 2008 9:01 am

NatalyaAF wrote:Taking trianglism to a whole new level...

nah, It's just that i like mathematics and physics.

and that brings me to the second point.
if you wanted to be even more accurate you could substract a number relative to the cosine of the angle to compensate for the friction of the ground. (my picture showed how to get the sine of the angle)
I'll try and work out an easier way to get this. (and work out some nice numbers for several kinds of surfaces)
but currently it involves setting the pole at a slight angle with steeper angles meaning more friction.
(leaning it in a direction that the force is smaller or the wrong direction, if negative or zero, no sliding occurs)
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Postby aybraus » Tue Dec 09, 2008 12:33 pm

*Bows in humility*
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Postby Rody » Thu Dec 11, 2008 1:53 pm

yeah, I have just finished my calculations.

if you want the friction  to have some effect then you should use:

a chain with length equal to:
Code: Select all
sqrt(8² + (8*f)²

(length in inches)

and the pole is angled towards the top of the hill with an angle equal to:
Code: Select all
arcsin( f /sqrt(1+f²)  OR arcsin( (8*f) /sqrt(8+(8*f)²)

where f is the coefficient of friction you are using.
(0 is no friction, larger numbers mean more friction, typically this is lower than 1)

for example a coefficient of 0.5 would give a chain length of about 9" and it would lean 27 degrees towards the hill.

you would then still take the shortest distance from the end of the chain to the pole.

also if the chain is on the side of the that would be leaning towards the top of the hill
(as it will when angles are small enough or it is standing on a flat surface)
then friction is larger than the force of gravity, and it won't slide.

if you decide not to construct the pole then you COULD use your calculator with the fomula:
Code: Select all
sqrt(8² + (8*f)² * sin(angle + arcsin( f /sqrt(1+f²))

negative numbers there mean no sliding occurs.
Last edited by Rody on Thu Dec 11, 2008 2:09 pm, edited 4 times in total.
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Postby Ross_Varn » Thu Dec 11, 2008 2:01 pm

So from that, can you calculate how much slide would happen?
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Postby Rody » Thu Dec 11, 2008 2:14 pm

yes, you can indeed. you use it calculate how much an object accelerates down a hill due to gravity.

basically I corrected an idea of aybraus for an easy way to find this.(see pic at top of page)

the post above (with the large amount of formulae) is me expanding on that idea and adding friction to it. you can ignore it if you want.
if not it is probably simplest to just use my precalculated example. which also desperately needs a diagram or picture for clarification)
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