anyone know any sites that give formulas for how gravity affects objects - for example a chain held at two points droops in a certain shape or a ball thrown horizontal, what path does this take - thanks

Well, the thing with the chain is, it depends on the radius of the links, the coarseness of the metal (friction) and the mass of the chain, however, normal chains fall n rope like formations, as far down as they can... Meaning, the closer you hold the two ends together, the less x,z tension there is. Even if you knew the "exact" shape, it would be very dificult to copy it onto a chain in any program... The ball question depends on Vi, and wind resistance. Without Wind resistance, which is very dificult to calculate, a ball will fall at the acceleration of gravity.. 9.8 meters per second. Meaning if the ball is thrown outward from 9.8 meters up it will hit the ground in 1 second, if it is not arched upwards. If you drop the ball from 9.8 meters it will hit the ground in one second. The main thing to remember is that gravity only effects the Y axis, forward velocity is not effected by gravity and therefore does not have to be conidered. Hope this helps.

Well, I think gravity pulls something @ 32 ft/sec. So a ball thrown horizontally falls at that rate. So does a bullet shot from a gun. So does anything regardless of how fast it's traveling horizontally. I don't know how to write that in equation form. A chain held at both ends drops in a parabola I believe. Again, no equation. But I'm fairly certain that it's a parabolic slope.

Right, 32 feet is 9.8 meters. And Chains/ropes/everything falls parabolically, unless it is being supported by it's own mass. Like a chain with Huge links, they would droop differently. It's better to use your own mind than equations when creating art, acctuality may interfere with the work... Most of the time it looks better if it is inaccurate, as people just expect reality, if all 3d art was realistic it would be boring, imo.

Attached Link: http://www.nas.com/~kunkel/hanging/hanging.htm

A chain or cable suspended freely from both ends forms a catenary curve, not a parabola. Don't worry, Galileo made the same mistake.(see link) The Gateway Arch in St. Louis was based on the catenary. Just to nitpik, the acceleration of gravity is 9.8 meters per second per second (it's seconds squared.) A ball thrown horizontally (or at any angle) will follow a parabolic curve if you neglect wind resistance. Johnpenn was right about the horizontal velocity being irrelevant. Suppose you have a gun lying horizontally on a table. A bullet is lying on the table near the edge. If the bullet rolls off the edge of the table at the exact moment the gun is fired (horizontally), which bullet hits the ground first? Answer: They both hit the ground at the same time. Steve S.

thanks very much sanvito this was more or less what i was looking for

Fz=M*G

Fz=gravitation in Newton (or in dutch zwaartekracht)
M=mass in Kilogramms
G=acceleration in m/s^2 (meter per sec^2) on earth the accelleration is 9.81 m/s^2

I Don't know the exact air/wind-resistence formula.
The energy of an object at a certain height is given with this formula:
Eh=MGh

Eh=heightenergy in Joule (and Joule is Newton per second)
G=gravitational accelleration (9.81 m/s^2)
h=Height in meter.

The energy of an object at a certain movement is given with this formula:
Ek=0.5mv^2

Ek=kinetic energy in Joule.
m=mass in kilogramm
v^2=(speed in meters per sec (m/s))^2

For throwing a ball horizontally you need these formulas:
Sx(t)=vt and Sy(t)=0.5g*t^2

Sx(t)=the length of the horizontal component of the objects movement over the time t. Here Sx= in meter and t= in sec.
V=speed of the object
t=time in sec

Sy(t)=the length of the vertical component of the objects movement over the time t. Here Sy= in meter and t= in sec.
G=9.81 m/s^2 (on earth)
t^2=(time in sec)^2

I hope this helps

thanks can i fit this into the attributes box

Anyone remember off hand the speed of terminal velocity?

Terminal velocity is a relative term. It is the speed at which wind resistance directly opposes acceleration. The speed would depend upon atmospheric density as well as the geometry of the object in question. Terminal velocity would therefore be much greater for a rock versus a feather. This is also the reason that a skydiver can slow themselves by spreading their arms and legs or increase their velocity by pulling their arms and legs in.

o.k. - do you remember the (maximum) terminal velocity of a solid iron sphere neglecting wind resistance,friction and ANY other extenal effects? There is an absoute answer to this question but I cant remember if its 188 feet per second or not.

opps - forgot to indicate that the question is in regards to planet Earth.

I'm not completely sure, but I think the terminal velocity of a skydiver with his/her arms and legs spread is something around 200 Km/h, but that could also be with the parachute already open.

I've get my final grades of this schoolyear next friday from my mentor, and he gives Physics. I'll ask him for the formulas nessecary to calculate the terminal velocity of an object, he might just know it.