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GT3
Neglecting air resistance and drive train resistance, for two cars of equal power but different weights would the top speeds be different? I think so.

There are at least two forces acting upon the inertia of a car at top speed. Weight by gravity and aerodynamics. It requires more force to hold a heavier car at a high top speed compared to a lighter car. Mass isn't irrelevant once there's enough force to push through it. Mass is a constant.

It requires more work/horsepower to get a heavier car at the same equal top speed.


Power (P) is work( W) done in unit time (t).

P = W/t

as work and energy (E) are same it follows that power is also energy consumed or generated per unit time.

Work is the product of force and the distance over which it moves. Imagine you are pushing a heavy box across the room.

Work = 1/2mv^2 - 1/2mv(0)^2

Power = [(1/2 mass x (velocity)^2) - (1/2 mass x (initial velocity)^2)] / Time

Disregarding aerodynamics:

100mph = 10mph^2
Initial velocity = 0.

[1/2 (1000kg)(100mph) - 0] / 1 hour = 50,000 units of power required to go 10mph for a 1000kg car.
[1/2 (1500kg)(100mph) - 0] / 1 hour = 75,000 units of power required to go 10mph for a 1500kg car.

50,000 = [1/2(1500kg)(x)] / 1 hour
50,000 = 750x / 1 hour
x = 66.7
Square root of x = Velocity
Square root of x = 8.1670068

So, if both cars put out the same amount of power, the lighter car would go 10mph. The 500kg heavier car reaches a top speed of 8.16mph.

Am I right?

Edit: I'm not talking about time to get to that speed, or the rate of acceleration, just the top speed at the end of the day.
MDT
Force is mass times acceleration. That means that the lighter mass will accelerate faster using the same force, i.e., same horsepower goes through the drivetrain the same for both. The little car will reach a higher top speed for any given time interval. Thats the short answer.

If you include lift, the heavier car may be able to go faster before lift is a concern. The little car (hare) will get there first, but may lose stablity before top speed, i.e., starts to float. The heavy car (tortose) will eventually catch up to the light car, since the hare might hesitate to go any faster and will stop trying to accelerate out of fear of losing control of the car. The tortose may pass him due to better stability created by its extra weight.

The hare may decide in the next race, to add a light weight air scoop, to use air to push the car down for more stability against lift. But that will add a force vector, opposite the direction of motion, that will increase with velocity. It won't change his mass much but will add a countering force. Under these conditions the hare will accelerate fastest near the beginning, i.e., little air force, and then gradually begin deceleration as the force of the air increases. The result will be a terminal velocity. By the end of the day, it will depend on when the tortoise's lift begins, since the tortoise will be still accelerating at the end.

Lsos
QUOTE (GT3+May 15 2007, 04:41 PM)
So, if both cars put out the same amount of power, the lighter car would go 10mph. The 500kg heavier car reaches a top speed of 8.16mph.

Am I right?


It might seem intuitively that you're right, but unfortunately, no. I didn't go through the equations, but I know somewhere you used them wrong because you disregarded drive train resistance and air resistance, and without these two factors, both cars should be able to go the same speeds. One might accelerate slower than the other, but both will reach the same terminal velocity...which, if we disregarded friction, there is none except the speed of light.

According to basic physics, you don't need power to travel at any velocity. A spaceship will keep going at whatever you sped it up to without any power. If you add power to it, it will simply accelerate. The reason cars do need power is to overcome air resistance and drive train losses. As such, a heavier car will have a theoretically lower top speed than a lighter car because of its greater drive train losses. However, this difference should decrease as speed increases, because at higher speeds the main factor that consumes power is air resistance, and drive train losses become negligible in comparison.
GT3
So if the two cars had exactly the same power, the same body shape, and everything else except for car A weighed more then car B they would both have the same top speed?

Thanks guys, I appreciate the help.

Also, did I go wrong in my calculation by assuming the same acceleration time for the car? I'm trying to figure out where I went wrong so I can rework the math!

I really do appreciate all the help from you guys!

-David
Lsos
QUOTE (GT3+May 19 2007, 02:58 PM)
So if the two cars had exactly the same power, the same body shape, and everything else except for car A weighed more then car B they would both have the same top speed?


They would have close to the same top speed. Car B would be a little faster due to less internal friction (rolling, bearing, etc). Rolling friction is the main force slowing a car down at low speeds, but air resistance grows exponentially with speed.

As for your calculations, they looked good to me except you were solving them for the the wrong question. What you solved for was how much power it takes to get a car up to 10mph in 1 hour.

In other words, if both cars had an engine with a power of 50,000 joules/hour, after 1 hour of accelerating the lighter car would be going 10mph and the heavier car 8.16mph (disregarding friction). However, that's not their top speed. As long as we continue applying the power, they would continue to accelerate indefinitely (again, disregarding friction).

If you want to solve for their top speed, the calculations are a bit more complex and require us to know many additional variables.

The simple equation is v = P/F

v: Velocity
P: Power (more power is more top speed)
F: Friction (more friction is less top speed)

On level ground, there will be two main frictional forces slowing a car down. Internal (in the bearings, wheels, etc), and air drag.

Friction due to air drag is expressed as Fd = Cd(pv^2/2)A

Where
Cd: Coefficient of drag (depends on shape of car. A teardrop-like shape generally has the lowest value)
p: Density of fluid (air will give less drag than water because it is less dense)
v: Velocity (Friction will rise exponentially depending on the velocity)
A: Frontal area of the car (the larger the area, the more drag)

Note that there is no mention of car weight in this entire formula.

Internal friction we will express as Fi, and I depends on the tire pressures, size, type, bearing material, lubrication, etc, etc.... and I will not even attempt to give a formula for this. However, generally this friction stays constant independent of velocity, but will rise with increased car weight.

Add these two components of friction and you get the total friction.

Top speed v = P/F

Note that friction due to drag depends on the velocity, and rises exponentially. So at low speeds the internal friction will be the main slowing factor, but quickly the air drag will overtake it. In fact, it takes 8x the power to overcome air drag at twice the speed. If my car which has 240 horsepower can go 150mph ....it will require 1920 horsepower to go 300mph.





bm1957
GT3, Lsos,

You have both made a slight error (could be typo, mindslip etc)

You speak of power being 'used' and 'amount of power'. This is not quite right. Power is simply a measure of how much energy is being used (converted) per unit of time. It tells you how quickly energy is being transferred from one form to another.

This doesn't affect what Lsos was saying as he included 'in 1 hour' but it makes a difference to the meaning of GT3s initial post.

As for the question, yes - if the air resistance/friction were the same for both cars at any given speed, the top speeds of both cars would be the same. It would be reached when the force provided by the engine is equal to the opposite force from the drag and friction effects. By definition of your question and my assumptions, both of these are equal for both cars.
Myrrdin
there would also be the difference in drag on the tires between a heavier and lighter car. (unless you included that in 'drive train resistance') it would have the same effect as slightly applying the brakes, since brakes use downforce to increase the friction on the tires to slow the car.

and also there would be a theoretical top speed measured at what point any vehicle did not have the power to increase the moment of the given weight. and that point would be reached for a heavier vehicle sooner than the lighter.

its the same law of physics that states the closer to the speed of light you get, the more energy required to gain the same amount of speed. ie going from 1mph to 2mph uses nearly no energy, while going from "speed of light -1mph" to "speed of light" requires an infinite amount of energy.
Lsos
QUOTE (Myrrdin+May 22 2007, 07:40 PM)

and also there would be a theoretical top speed measured at what point any vehicle did not have the power to increase the moment of the given weight. and that point would be reached for a heavier vehicle sooner than the lighter.

its the same law of physics that states the closer to the speed of light you get, the more energy required to gain the same amount of speed. ie going from 1mph to 2mph uses nearly no energy, while going from "speed of light -1mph" to "speed of light" requires an infinite amount of energy.

I sure did think especially of the tires when I mentioned drive train resistance.

As for the top speed...it would be reached when the force supplied by the engine is countered equally by the force of friction. Not before, not after. Whether a car weighs 1000lbs or 10000000lbs, the formula F=ma still holds true. And unless Force = 0, and until acceleration ceases, the car hasn't reached its top speed.

In a frictionless environment, this speed would be approaching light speed. This is true no matter how powerful the engine is and how much the vehicle weighs.
Myrrdin
QUOTE (Lsos+May 23 2007, 07:15 PM)
I sure did think especially of the tires when I mentioned drive train resistance.

In a frictionless environment, this speed would be approaching light speed. This is true no matter how powerful the engine is and how much the vehicle weighs.

apologies for not being specific. my comment about tires was for the guy who originally posted the question.

and as far as it reaching the speed of light in a frictionless environment, scientists have maintained for years that even in the vacuum of space trying to reach near that speed would take more energy than the whole planet can produce. unless i didnt understand your reply
Lsos
QUOTE (Myrrdin+May 23 2007, 10:33 PM)
and as far as it reaching the speed of light in a frictionless environment, scientists have maintained for years that even in the vacuum of space trying to reach near that speed would take more energy than the whole planet can produce. unless i didnt understand your reply

Right, I agree. I was just making an example...saying that as long as there's a force acting on the vehicle, no matter how small and no matter how much the vehicle weighs, it will still continue to accelerate and thus hasn't yet reached its top speed. Its only speed limit would be the speed of light.

How to keep applying that force and where to get energy for it is beyond our current technology and understanding of physics, and is a topic for another discussion.
GT3
Thanks guys, I really, really appreciate all the help.
Nick71
Very useful discussion indeed..
Sorry for bringing back this subject after so long, but I was wondering the following:

Say you have a car travel ling at its top speed of 130mph. At that speed, what percentage of the resisting force is due to air resistance, and what percentage is due to internal friction (transmission, kinetic friction of tires, etc)?

What I'm trying to understand is what effect would an increase of the weight of the car by, say, 100kg have on its top speed. I thick the answer is "very little" (1-2 mph??...) if the top speed is as high as 120-130mph (assuming, of course, that the tires are properly inflated for this weight increase)... But I'm trying to put some Physics behind this impression I have....

Thanks & apologies for my not so good English (I'm from Greece).

Nick
Noumenon
Your correct weight plays little difference in terminal velocity,.. it just takes a longer distance to achieve top speed for a heavier vehicle with the same power.

The force which will be balanced by the cars peak horse-power (gearing setup correctly) is,...

F = 1/2*CdA*p*v^2 + CrW ,... where,

Cd = Coefficient of drag for the particular vehicle. It is measured in a wind tunnel.
Cr = Coefficient of rolling resistance for vehicle.
p = Density of air,.. A = Cross sectional area of the car,...W = Weight of the car,...v = Velocity.

As you can see the first term which involves air drag, increases as the square of the velocity, while the second term involving weight is independent of velocity.

------------
Incidentally, the bhp required to go a certain velocity is,... bhp = F*v/550

Also since air drag is the largest factor here, the condition of that air makes a significant difference, probably more so than the weight at high speed. The air density above (p) can be estimated as follows;

p = P/(T*R) ,... where,

P = air pressure
T = temperature
R = gas constant (1718 ft-lb/slug)

------------
As an example, my Suzuki Hayabusa has a drag Coefficient of 0.561 (a good car will have a lower one than a motorcycle) and a cross sectional area of only ~6f^2, so if its 70*f outside at normal barometer pressure 29.92 psi, the air-drag force against the bike at 190mph is around 320lbs (probably a little rear wheel slippage at that speed). If its colder the air is denser; a little more bhp will result, but small compared with the increase in drag force.
Catch
So to conclude this thread can we safely say, air resistance being equal.

Car a) 100 WHP car, 1000 lbs. - Power to weight = .1hp/lb.
Car cool.gif 100 WHP car, 500 lbs. - Power to weight = .2hp/lb.

Car 'B' will only be slightly faster than car 'A', (talking about terminal velocity) and this, only because of the added friction from the drag on the tires. (Pretend both cars have the same drag from aero bits to keep the cars on the ground @ around 170 mph which are both cars' terminal velocity.)

Thanks!
light in the tunnel
This thread evokes a question about bicycling:

If two bicycles have the same wind-resistance profile and tire-contact with the road, why would a heavier bicycle-rider combination require more power to maintain a certain speed than another?

Shouldn't maintaining 15mph with 250lbs use the same amount of power as the same speed at 150lbs? This is assuming that the two cases have the same loss of momentum due to air-friction? Does the weight difference translate into a proportional difference in rolling resistance even if the contact patch of the tires is kept constant?
flyingbuttressman
QUOTE (light in the tunnel+Feb 2 2010, 09:06 AM)
This thread evokes a question about bicycling:

If two bicycles have the same wind-resistance profile and tire-contact with the road, why would a heavier bicycle-rider combination require more power to maintain a certain speed than another?

Shouldn't maintaining 15mph with 250lbs use the same amount of power as the same speed at 150lbs? This is assuming that the two cases have the same loss of momentum due to air-friction? Does the weight difference translate into a proportional difference in rolling resistance even if the contact patch of the tires is kept constant?

"Faster" implies more than just top speed. The lighter bike/rider combo is less resistant to changes in velocity and direction. Semi trucks can have 12 cyl/500HP engines, just like some high-end sports cars, but they won't be able to touch a sports car's performance.
light in the tunnel
QUOTE (flyingbuttressman+Feb 2 2010, 02:19 PM)
"Faster" implies more than just top speed. The lighter bike/rider combo is less resistant to changes in velocity and direction. Semi trucks can have 12 cyl/500HP engines, just like some high-end sports cars, but they won't be able to touch a sports car's performance.

Granted the heavier bicycle+rider will require more energy and force to accelerate or decelerate. But it is probably counterintuitive to some cyclists that maintaining constant speed would not require more energy for the heavier bicycle+rider than the lighter one given the same contact-patch between tires and road.

The thing I'm wondering about is whether just the additional pressure on the same contact-patch would increase rolling resistance significantly, esp at 15+mph. Usually I think of wind-resistance being more important starting at around 15mph, but I'm not sure how to figure out the effect of weight on rolling-resistance of the tires.

Same principle with Semi Trucks. Intuitively, their larger weight would seem to demand more energy to maintain constant speed than for a passenger vehicle. However, maybe it is only the additional tires and road-contact that reduce energy/fuel efficiency, along with wind resistance perhaps since they don't look as streamlined as many passenger vehicles.
flyingbuttressman
QUOTE (light in the tunnel+Feb 2 2010, 11:27 AM)
Intuitively, their larger weight would seem to demand more energy to maintain constant speed than for a passenger vehicle.

If that's what you intuition tells you, your intuition is broke.

Tell me, which is heavier, a ton of lead or a ton of feathers?
adoucette
QUOTE (light in the tunnel+Feb 2 2010, 11:27 AM)
I'm not sure how to figure out the effect of weight on rolling-resistance of the tires.

light in the tunnel
QUOTE (flyingbuttressman+Feb 2 2010, 05:08 PM)
If that's what you intuition tells you, your intuition is broke.

Tell me, which is heavier, a ton of lead or a ton of feathers?

I agree with your point. I think it's striking how much cyclists invest in lighter bikes and parts, though, if maintenance of constant speed requires about the same amount of energy either way. It sounds like what they should really be investing in is harder tires with less elastic tread.
light in the tunnel
QUOTE (adoucette+Feb 2 2010, 05:34 PM)
http://www.tut.fi/plastics/tyreschool/modu...xt_1/3/3_3.html

Arthur

Interesting article, Thanks.

It reminded me of an idea that once occurred to me to design tires/wheels that expand a narrow "skate" in the center of the tire that would prevent contact of the rest of the tire at high speed while traveling in a relatively straight line.

Of course, such tires would have to immediately resume full contact with the road surface for emergency maneuvering. But on long highway trips, how much fuel efficiency would be gained by rolling on a narrow central "skate" wheel/tire? Likewise, how much tread wear would be prevented?
flyingbuttressman
QUOTE (light in the tunnel+Feb 2 2010, 02:24 PM)
I agree with your point. I think it's striking how much cyclists invest in lighter bikes and parts, though, if maintenance of constant speed requires about the same amount of energy either way. It sounds like what they should really be investing in is harder tires with less elastic tread.

Wow, missing the point again.

Lighter bikes = faster acceleration and quicker stopping. Also, don't forget that the "same energy at constant speed" thing only applies to level terrain. On a hill, a lighter vehicle is much easier to move.
light in the tunnel
QUOTE (flyingbuttressman+Feb 2 2010, 07:37 PM)
Wow, missing the point again.

Lighter bikes = faster acceleration and quicker stopping. Also, don't forget that the "same energy at constant speed" thing only applies to level terrain. On a hill, a lighter vehicle is much easier to move.

. . . because gravity is a force, i.e. F=MA, so overcoming the same deceleration going uphill requires more force for a greater mass.

I was only thinking of level ground, and I still think there are cyclists who would find it counterintuitive that a heavier bike requires the same force to maintain the same speed on level ground.

I'm not arguing that less weight doesn't have other benefits.
NoCleverName
Forgetting aero drag and just focusing on rolling resistance, equal power output results in a lower speed for the heavier bike and rider. This is because rolling resistance is due to static friction which is proportional to the normal force (i.e., weight). This force must be overcome. It takes work to overcome force. Power is work per unit time. Velocity is distance per unit time. Put together, v = P/F. Clearly, with less resistive force and P constant, v must increase.

Countering the increase in speed of the light bike is the fact that aero drag increases as the cube of speed. So, while the lighter biker has the ability to go faster, he can't reach his ultimate advantage over the heavier biker due to drag ... but he still goes faster for the same power output.
light in the tunnel
QUOTE (NoCleverName+Feb 3 2010, 12:20 AM)
Forgetting aero drag and just focusing on rolling resistance, equal power output results in a lower speed for the heavier bike and rider. This is because rolling resistance is due to static friction which is proportional to the normal force (i.e., weight). This force must be overcome. It takes work to overcome force. Power is work per unit time. Velocity is distance per unit time. Put together, v = P/F. Clearly, with less resistive force and P constant, v must increase.

Countering the increase in speed of the light bike is the fact that aero drag increases as the cube of speed. So, while the lighter biker has the ability to go faster, he can't reach his ultimate advantage over the heavier biker due to drag ... but he still goes faster for the same power output.

Interesting, but what is the influential factor in rolling resistance? You say it is proportional to weight, but it is not due to intertia, since inertia is only resistance to changing velocity, i.e. acceleration or deceleration. Therefore I assume it must be due to wheel and tire flexing, and the friction generated by the contact-patch of the tire against the road.

So a less flexible wheel with a harder (inflated) tire, and harder tread (worse for traction/maneuvering) would effectively neutralize weight as factor in the amount of energy required to maintain a constant cruising speed, no?

Really it doesn't matter much anyway because of the exponential increase in wind-resistance becoming a more important factor than rolling resistance @>15mph. Still, it is somewhat ironic that cyclists fixate on reducing equipment weight if it doesn't make a difference in constant cruising on flat ground. I suppose that getting up hills, accelerating, and braking are frequent and important enough issues for many cyclists that a lighter bike is worth the trouble and cost.
boit
QUOTE (light in the tunnel+Feb 2 2010, 02:06 PM)
This thread evokes a question about bicycling:

If two bicycles have the same wind-resistance profile and tire-contact with the road, why would a heavier bicycle-rider combination require more power to maintain a certain speed than another?

Shouldn't maintaining 15mph with 250lbs use the same amount of power as the same speed at 150lbs? This is assuming that the two cases have the same loss of momentum due to air-friction? Does the weight difference translate into a proportional difference in rolling resistance even if the contact patch of the tires is kept constant?

I honestly didn't know that is so (Heavier bicycle-rider combination require more power to maintain certain speed. What I know is friction is mass dependent therefore loss to friction can't be ruled out, especially at cycling speeds.
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