Been thinking at work (when I'm not busy) and I figured one of you could answer this..



If there are 2 cars, A and B, about to drag race, who will win? (data below)



a. 300rwhp 3000lbs.



b. 150rwhp 1500lbs.



Now assuming both cars have identical gearing and the drivers are identical in performance...WHO WINS?! or does anyone!?



Sean

 

how about the torque... is the torque identical to the hp?

 

never even thought about it...so I'll say...sure, identical to hp..



Sean

 

You'd really have to look at their momentum in a scenario like this one.



momentum = mass x velocity [p=mv]

impulse = mass x (velocity final - velocity intial) [p=m(v2-v1)]

inertia is momentum at rest...

1hp = 550ft(lbs)/s of Force

1hp = 746 watts of energy



Theoretically they'd be exactly the same in acceleration if I'm not mistaken.

 

the only thing i can think of, is that the 3K car would lose less momentum when its shifting, its a pretty god damn small increment, but thats the only thing i can see in terms of a car winning.

 

thanx ppl, that's what i was looking for, i was wondering if the cars were halfed exactly if they would still remain as fast........So would torque really affect this scenario as the smaller car would not need as much, and the bigger would, but the torque would still be the same as the hp and thus having the same effect on both cars... so they would still be =..right?

 

In real life there are going to be impedances like friction, however on a theoretical frictionless plain, they'd still be exactly the same.



Crash... What you're looking at is having to over come intertia. Both have the same power/weight ratio so they'd both overcome inertia at the same rate. However the 3000lb car would have double the momentum as the 1500lb at the same speed.



Torque would play into the equation if you factored it in. Torque is basically rotational force. By definition: The measure of a force's tendency to produce torsion and rotation about an axis. Think of a door. If you push a door real close to the hinges, it's much more difficult than it is to push it where the handles are located.



T = F*D where T = Torque, F = Force, and D = distance as seen below.

 

Wow, thanx Exyom, keep it coming, i like learning new stuff....i knew the basics, just not the terms or all the rules applying to the problem...bah, I have to go to work



Sean

 

Force=Mass*Acceleration



Negating friction and air resistance, If the engines have exactly proportional torque curves, and the same size tires, then they will go the same speed. However, friction and air resistance are relatively constant so the same force on a larger mass will produce a smaller acceleration. So the heavier car will be faster due to the drag having less of an effect.