What is the equation for deacceleration of a moving object?
I'm mostly looking for taking into account loss of force over distance and air friction in a standard atmosphere. Weight isn't much of an issue.
~Polatrite~
P.S. Let me know if I'm off-base or need to explain things a bit more.
ID:153469
 Jan 8 2004, 4:21 pm
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I thought terminal velocity was a constant; it's the point at which it is reached that is variable. Of course, I could be wrong =P
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I thought terminal velocity was a constant; it's the point at which it is reached that is variable. Of course, I could be wrong =P The force of air resistance depends on mass and cross sectional area. This force increases with the force you apply on it. So eventually the force of gravity matches that of the force of air resistance canceling each other out leaving you at a constant speed. But if you change either your mass or shape you change the force due to air resistance and eventually it again levels out with gravity only at a new speed. [edit] I could be wrong though since I'm posting this from memory so you probably want to check a physics book for better information if it's needed :). | |
sapphiremagus wrote:
I thought terminal velocity was a constant; it's the point at which it is reached that is variable. Of course, I could be wrong =P And are! The terminal velocity of an object depends on its aerodynamic properties, and how they change as the object rotates or deforms as it falls. The atmosphere itself matters, too; obviously terminal velocity would be lower in a thicker atmosphere because of greater drag. This is why parachutes work; it's not that they make it take longer for you to reach terminal velocity, but that they alter your terminal velocity to a speed that won't kill or injure you on impact. Lummox JR | |
I'm mostly looking for taking into account loss of force over distance Force is never lost over distance. =) An object which is accelerated in space will never stop until something pushes on it from the opposite direction. An object in Earth's atmosphere drops due to gravity, not due to a loss in momentum (though, of course, air resistance does play a factor here). If you drop a ball vertically and project another horizontally, both balls will hit the ground at the same time. In any of these cases, deceleration is extremely complicated; just about any game you play on the market doesn't use real physics to calculate air resistance on any object, primarily because it requires complex geometry, and secondarily because it's unnecessarily CPU-intensive. If you still want to have reasonably accurate air resistance, then you can use a basic formula -- heavier objects can be abstracted to have more surface area (this isn't true, but works well enough as an abstraction), and objects can have a floating point value for aerodynamics which determine negative acceleration due to air resistance. You'll really have to develop formulas on your own... but that should give you a baseline to start from. | |
Heh I'd go with just comming up with some reasonable estimate. For proper air resistance calculations you need to know the mass and the cross-sectional area that's hitting the resistance.
The force of air resistance is proportional to force pushing against it. This is why when falling objects reach a terminal velocity(ie they stop accelerating and reach a constant speed) which varies depending on shape and mass.
So what do you need this for :P?