Recent content by ivtec259

Oh so you mean that saying for example that "My car weighs 1500 kg" is wrong and you should say "My car's mass is 1500kg"? And by the way, I do not live in the US.

Where are they teaching you this? Just curious. Because last I checked with my physics instructor, a kilogram is a unit of mass and nothing else. If you want to talk about forces, then talk Newton. And how can you believe that I meant a force, when I wrote m = 1500 kg. I did not write F = 1500...

Whaaaaat. No. I think you are the one that needs to check again and then recheck, kind of like in Transformers 1 when that helicopter comes back from the dead. 3,307 pounds scale weight (lbf) is a force. If you want to convert to kilograms, you first need to convert you force into a mass. But...

Sure, not a problem. Let m = 1500 kg, a = 10 m/s^2, P = 535 hp = 393000 W. Then velocity (v1) = 26.2 m/s, position (x1) = 34.2 m and time (t1) = 2.62 s. The power for stage 2 is the constant 535 hp, so I'm not sure what you mean by "calculate a similar power for stage 2". If you want to...

You are absolutely correct. But P and a are never constant at the same time in that formula. The formula is only valid for t>t1 (stage 2) where P is constant and a is not. Maybe it is more pedagogic to name a amax and P Pmax instead. The acceleration value in the formula is not the acceleration...

That would be a good idea, thanks for the velocity formula.

I seem to be lost in the imperial system. I am going to stick to the metric system then, and those who want to can convert to other units if they want to. I just thought it would be polite to show the mass in lbm or whatever is the right unit for mass.

No it is not. a is assumed to be 10 in this case, which corresponds to about 1g of acceleration. This is reasonable for normal car tires. I thought I was clear that I am talking about mass, not force. m = mass. lbs is a unit of mass and lbf is a unit of force right? Anyway 3307 lbs is 1500 kg...

I am sorry if i was unclear in that post. I will try to make it more clear. This is not constant acceleration the whole quarter mile, only until the power of the engine starts to be the limiting factor. a is the maximum acceleration limited by traction of the tires. 3307 US pounds is definitely...

Well there is only one P and it is a constant, so you can call it whatever you want but Pmax is more suitable in this case I think, because the power will never be more than this. I did some algebra and the best I could do was this: ##x(P,m,t,a) = \frac{m}{3P} \left( \frac{P}{m} \left(2t-...

Yes I think so, and hopefully the expression will become more pretty once you simplify it.

But t1 is not an independent variable since ##t_1 = \frac{P_{max}}{ma_{max}^2}##

Thank you for that! I will try to do it myself from there.

From what I understand, the traction is only limited at lower speeds. At higher speeds the power of the engine is too low to create wheel-spin in a 10s car, hence the 2-stage acceleration that i talked about. But sure in top fuel dragsters i can imagine there is a case for aerodynamic down force...

Yes but how do i continue from there? I already derived that equation, and it only works when the initial velocity is zero right?