lets say we have to cars one that weights
2000lbs with 200hp and 100ft/lbs torque
3000lbs with 300hp and 150ft/lbs torque

which would give them 1hp for every 10lbs
assuming both cars are the exact same shape and same gearing would one be faster then the other?

My brain says they should be exatly the same but for some reason im thinkin they won't be?

 

The answer is no because the aerodynamic drag will be the same in both cases. If you look at Newton's second law, i.e. F = ma, where F is the sum of the forces acting on the body (car). In a straight line (no cornering forces), the sum of the forces is equal to the force transmitted through the contact patch from the engine, the aerodynmic drag, the tire rolling resistance, and any rotational inertia. To simplify, neglect the tire rolling resistance and the rotational inertia, and you are left with drag. The drag does not depend on the engine power or increase in mass per your initial considerations.

F = ma

Fw - Drag = ma

Let a1 = acceleration of car 1 and mass 1
Fw1 = Tire tractive force from engine for config. 1.

Let a2 = 2Xa1
Fw1 = 2XFw2

a1 (1 - D/Fw1)
-- = --------------
a2 (1-D/2*Fw1)

The denominator is always a larger number than the numerator unless D = 0, or when the speed is zero (i.e. from a standing start). Therefore, the heavier car will accelerate faster at all speeds (except at the initial start up where the acceleration of the heavier car equals that of the smaller car.

In a corner, the same type of analysis will show the lighter car will have an advantage, because of tire traction nonlinearity (i.e. a tire generates increasing lateral grip at a decreasing rate with increase vertical tire load).

Hope this helps.

 

but wouldn't it bw just as hard for both cars to overcome the drag?

because the have the exact same power to weight ratios?

i just can't see why the heavier car would accelerate faster (sorry if im dumb trying to understand)

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by ekim952522000 &raquo;</TD></TR><TR><TD CLASS="quote">but wouldn't it bw just as hard for both cars to overcome the drag?

because the have the exact same power to weight ratios?

i just can't see why the heavier car would accelerate faster (sorry if im dumb trying to understand)</TD></TR></TABLE>

It's because, there is more power and torque per unit mass for the heavier more powerful car. Hence, this extra power to weight/torque can be used for acceleration.

Look at it this way with some numbers:

Fw - D = ma

Let m = 1, let Fw = 100, and let D =50

Fw - D = 100 -50 = 50 = 1a

a = 50 (units)

Now, double the mass and double the Fw and leave D alone

200 - 50 =150 = 2a

a =75 (consistent units)

So, the heavier and more powerful car has more acceleration.

 

Depends how you define "faster."

If you are talking ultimate, straightline speed the more powerful of the two will get it done. Ultimately, more poop is going to push the same shape through the air faster, regardless of its mass. But note that because drag increases with the square of the velocity, the difference might not be as great as one would expect.

If you are concerned with lap times? It might be a different story in the real world. Race cars spend relatively little time at maximum velocity, so mass - in terms of accelerating it positively (on the gas), negatively (on the brakes), and laterally (in the corners) becomes a relatively greater part of the equation.

If you are worried about finishing position at the end of a race of some duration, that's yet another issue. A powerful, heavy car might not have brakes at the end of a race and will be harder on its tires, perhaps costing enough in lap time to allow a lighter, easier-on-equipment car to beat up on it over a race distance.

K

 

Take a Bentley Continental GT and Lotus Elise for example. Both have similar power/weight ratios, and (of course they have different gearing, drivetrain configurations, etc., but this is just an example) they accelerate to 60 in about the same time. However, past a certain speed, aerodynamic drag begins to have more of an effect than weight. This is why the huge, very powerful Bentley does 200mph while the relatively low-powered but also lightweight Elise would struggle to hit 150mph.

Side note: your sample cars would have to rev to at least 9500rpm in order to make that much hp out of those torque numbers.....

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by ekim952522000 &raquo;</TD></TR><TR><TD CLASS="quote">i just can't see why the heavier car would accelerate faster</TD></TR></TABLE>
If there would be 0 drag and assuming all other friction forces to be zero, indeed both would accelerate equally as fast since only F = ma will apply to accelerate the mass of the car. If "F" and "m" remain proportional to each other, "a" - the acceleration of the car - will be constant. Acceleration would also be linear, as in both cars would accelerate from 20-50 mph as fast as they would from 120-150 mph. Once you introduce drag, the heavier, more powerful car will have the advantage. The force of the drag will be larger relative to the force put out by weaker motor, and smaller in relation to the force put out by the stronger motor. Therefore, the more drag is introduced in the equation, the more it will have an impact against the car will less power than it will have on the car with more power. For this reason the Bentley vs Elise example holds true, and also the reason that at high enough speeds, a heavier car with a more powerful engine may still yet accelerate faster than a lighter car that even has a more favorable power/weight ratio.

If you want to consider on track overall performance, then it gets a lot more complex as Knestis suggested.

 

whats the area under the power curve for each?

 

since its the road racing forum and not the drag racing forum the lighter car will be faster, since tire grip doesn't increase exponentially with weight.

Comparing the bentley gt and lotus are useless, the lotus has like 1/6 the tq depending on the engine and the profile of the cars makes it somewhat less stable, on top of which I doubt even in the 111r the celicas geared over 150 anyway

and the elise has a cd of .36 vs the bentleys .32 which can have a bit of an impact at those speeds

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by suavacito &raquo;</TD></TR><TR><TD CLASS="quote">
Comparing the bentley gt and lotus are useless, the lotus has like 1/6 the tq depending on the engine and the profile of the cars makes it somewhat less stable, on top of which I doubt even in the 111r the celicas geared over 150 anyway

and the elise has a cd of .36 vs the bentleys .32 which can have a bit of an impact at those speeds</TD></TR></TABLE>
Okay, it's not useless.... the Lotus has 1/4 the torque, not 1/6. A 0.04 cD difference is significant but not enough to make a 50mph difference. It was a SIMPLE illustration as I stated. Make the Lotus as slippery as an Insight with torque that is "only" 1/3 of the Bentley's (as the hp is) and give it whatever gearing you want... it won't get anywhere close to 200mph and that's the point.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by White98LS &raquo;</TD></TR><TR><TD CLASS="quote">
Okay, it's not useless.... the Lotus has 1/4 the torque, not 1/6. A 0.04 cD difference is significant but not enough to make a 50mph difference. It was a SIMPLE illustration as I stated. Make the Lotus as slippery as an Insight with torque that is "only" 1/3 of the Bentley's (as the hp is) and give it whatever gearing you want... it won't get anywhere close to 200mph and that's the point.</TD></TR></TABLE>

It's not Cd that's important, it's Cd*A that's important. Got to compare apples to apples when discussing forces. Cd's only compare the efficiency of the shape. And the Lotus Cd*A is significantly smaller than a Bentleys even if their Cd's are similar.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Johnny Mac &raquo;</TD></TR><TR><TD CLASS="quote">

It's not Cd that's important, it's Cd*A that's important. Got to compare apples to apples when discussing forces. Cd's only compare the efficiency of the shape. And the Lotus Cd*A is significantly smaller than a Bentleys even if their Cd's are similar.</TD></TR></TABLE>

i am glad you said this as it is often left out.

now riddle me this batman, if you have two identical cars with identical engines etc. and you add ballast to one of them you are saying that the heavier car would accelerate faster after they reach a speed where drag is significant enough to matter?

 

that is what i dont get.

It souns like youre saying power has more to do with top speed than weight

How about a
3000lb car with 300hp and 150FT/lbs of torque
2000lb car with 300hp and 150ft/lbs of torque
everything else is the same

According to your theroy's the should have almost the same top speed?

because power matters more than weight?

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Mr.E.G. &raquo;</TD></TR><TR><TD CLASS="quote">

i am glad you said this as it is often left out.

now riddle me this batman, if you have two identical cars with identical engines etc. and you add ballast to one of them you are saying that the heavier car would accelerate faster after they reach a speed where drag is significant enough to matter?

</TD></TR></TABLE>

I don't understand your question: Two identical cars, identical engines, but one has more ballast? I think you meant to say something different.

 

J Mac get back at wrenching on that CRX our race is in less than 3 weeks.
My car is finally running since the November teardown.


Edo

 

I am not sure why this is so hard to understand.

Putting all other friction forces aside, the two main factors impeding a car's acceleration are a) the force required to accelerate the mass of the car (F = mass * acceleration) and b) the force required to work against drag (D), which is the product of Cd * frontal area, as Johnny Mac pointed out. At a speed of 0 mph where D = 0, all the engine's power will be used to accelerate the car's mass (F = ma). Keeping engine power constant, it's clear that the more mass it has to move, the slower it will be able to accelerate it.

The faster you go, the greater the force of drag, D, will be. The greater D becomes, the less force is left to accelerate the car (F = ma). A car's top speed occurs when all the power of the engine is used to work solely against D. As a result F = 0 = ma and so acceleration = a = 0. There is no force left that can further accelerate the car, and so it will continue to travel at a constant speed, its drag limited top speed.

As an example:

Toy car A's power is able to exert 12 N (N = Newton = kg*m/s^2) of force. It has a mass of 6 kg.
Toy car B's power is able to exert 6 N of force. It has a mass of 3 kg.
Both cars A and B can be said to have identical “power to weight” ratios of 2, and given F = ma, both cars would accelerate just as fast in a 0 drag environment.

Now you introduce the force of drag, D. Assume drag exerts a force on each car of 3 N.

Car A now has 9 Ns of force remaining that can be used to accelerate it. Its acceleration can be found by F = ma -&gt; 9 = 6a. Acceleration = a = 9/6 = 1.5 m/s^2.

Car B has 3 Ns of force remaining that can be used to accelerate it. Its acceleration can be found by F = ma -&gt; 3 = 3a. Acceleration = a = 3/3 = 1 m/s^2.

As it can be seen, although both cars A and B have identical power to weight ratios, once drag is introduced, car A will have a better acceleration, given everything else equal of course. By the same token, it's easy to see now why car A would also have a higher top speed than car B.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by ekim952522000 &raquo;</TD></TR><TR><TD CLASS="quote">that is what i dont get.

It souns like youre saying power has more to do with top speed than weight

How about a
3000lb car with 300hp and 150FT/lbs of torque
2000lb car with 300hp and 150ft/lbs of torque
everything else is the same

According to your theroy's the should have almost the same top speed?</TD></TR></TABLE>
Yes.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by ekim952522000 &raquo;</TD></TR><TR><TD CLASS="quote">that is what i dont get.

It souns like youre saying power has more to do with top speed than weight

How about a
3000lb car with 300hp and 150FT/lbs of torque
2000lb car with 300hp and 150ft/lbs of torque
everything else is the same

According to your theroy's the should have almost the same top speed?

because power matters more than weight?</TD></TR></TABLE>

I think you meant to say "top speed has more to do with power than with weight".

The above scenario is true, both will acheive the same top speed. The heavier car will simply take longer to get there...you still have to accelerate the mass.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Hracer &raquo;</TD></TR><TR><TD CLASS="quote">I am not sure why this is so hard to understand.

Putting all other friction forces aside, the two main factors impeding a car's acceleration are a) the force required to accelerate the mass of the car (F = mass * acceleration) and b) the force required to work against drag (D), which is the product of Cd * frontal area, as Johnny Mac pointed out. At a speed of 0 mph where D = 0, all the engine's power will be used to accelerate the car's mass (F = ma). Keeping engine power constant, it's clear that the more mass it has to move, the slower it will be able to accelerate it.

The faster you go, the greater the force of drag, D, will be. The greater D becomes, the less force is left to accelerate the car (F = ma). A car's top speed occurs when all the power of the engine is used to work solely against D. As a result F = 0 = ma and so acceleration = a = 0. There is no force left that can further accelerate the car, and so it will continue to travel at a constant speed, its drag limited top speed.

As an example:

Toy car A's power is able to exert 12 N (N = Newton = kg*m/s^2) of force. It has a mass of 6 kg.
Toy car B's power is able to exert 6 N of force. It has a mass of 3 kg.
Both cars A and B can be said to have identical “power to weight” ratios of 2, and given F = ma, both cars would accelerate just as fast in a 0 drag environment.

Now you introduce the force of drag, D. Assume drag exerts a force on each car of 3 N.

Car A now has 9 Ns of force remaining that can be used to accelerate it. Its acceleration can be found by F = ma -&gt; 9 = 6a. Acceleration = a = 9/6 = 1.5 m/s^2.

Car B has 3 Ns of force remaining that can be used to accelerate it. Its acceleration can be found by F = ma -&gt; 3 = 3a. Acceleration = a = 3/3 = 1 m/s^2.

As it can be seen, although both cars A and B have identical power to weight ratios, once drag is introduced, car A will have a better acceleration, given everything else equal of course. By the same token, it's easy to see now why car A would also have a higher top speed than car B.</TD></TR></TABLE>

ok man that was a great read now i get it

 

For those of you who might like numbers to stuff in your pipe, I simulated the following scenario using LapSim for straight-line acceleration:

Car 1:"Exemple streetlegal sports car", as is, supplied with software
mass = 1500 kg
torque = 385 Nm max
HP = 380 PS max
Cd = 0.31
Frontal area = 1.81 m2
Longitudinal tire slip = 10%

Car 2:"Exemple streetlegal sports car", but half the mass, half the torque, same power to weight ratio and drag
mass = 750 kg
torque = 192.5 Nm max
HP = 190 PS max
Cd = 0.31
Frontal area = 1.81 m2
Longitudinal tire slip = 10%

<U>Results</U>

Car 1
0-60 km/h: 2.28 s
0-100 km/h: 4.46 s
0-160 km/h: 9.46 s
0-200 km/h: 14.88 s
Top speed (drag limited): 303 km/h

0-100 m:5.50 s
0-400 m:12.56 s
0-1000 m: 22.66 s

Car 2
0-60 km/h: 2.32 s (diff -0.04 s)
0-100 km/h: 4.66 s (diff -0.20 s)
0-160 km/h: 10.46 s (diff -1.00 s)
0-200 km/h: 18.26 s (diff -3.38 s)
Top speed (drag limited): 243 km/h (diff -60 km/h)

0-100 m:5.58 s (diff -0.08 s)
0-400 m:12.90 s (diff -0.34 s)
0-1000 m: 23.76 s (diff -1.10 s)


As you can see, the initial times to 60 km/h are very close, but the faster one goes, the more the low power car is hindered by air drag and the more it loses to the big, heavy, high power car.

And for what really counts, the lap-time around Mosport is:

Car 1: 1:37.36

Car 2: 1:36.15 (faster by 1.21 seconds)

So while the heavy powerful car might win in the standing kilometer, the lighter less powerful car would win the race. It corners and brakes faster, and makes generally better use of its tires. Colin Chapman figured that out a rather long time ago: Add Lightness

 

that was some cool info