View Poll Results: Who won the debate: MK (HP) or VR (Torque)
Mk won with a simple to understand concept that HP determines torque at the wheels at any speed.
25
17.48%
MK won: When comparing equal HP cars, the one with less torque COULD be better on the road course.
6
4.20%
VR won: When comparing equal HP cars, the one with more torque is better on a road course.
44
30.77%
Neither, as physics dont apply to race cars
18
12.59%
I don't want to open this can of worms again!
50
34.97%
Voters: 143. You may not vote on this poll
Poll: Who won the HP vs Torque debate?
03-23-2009, 06:41 PM
#421
Rennlist Member
Thread Starter
Where did you get this formula from? Hp*t = KE? Hp-seconds is a unit measure of work, not a rate of change of kinetic energy.
OK, this is easy enough to prove; this data is readily available. Let's assume an 09 C4S, 385hp, 3400lbs. (287kW, 1542kg). I think we're all familiar with cars in this general performance category 
Let's ignore torque, ignore drag (only a small factor at these speeds with this car) and simply compute by kinetic energy:
HP*t=Ek=1/2mv^2
.
Let's ignore torque, ignore drag (only a small factor at these speeds with this car) and simply compute by kinetic energy:HP*t=Ek=1/2mv^2
.
03-23-2009, 06:58 PM
#422
The Penguin King
Rennlist Member
Rennlist Member
03-23-2009, 07:18 PM
#423
Three Wheelin'
Bill,
I think this is where you need to alter your formula.
F = ma gives you the acceleration at an instant moment. Instant moment is when you get the F pulse from the engine. In other words when a cylinder fires.
To calculate the acceleration of the car you need acceleration over a time-period. It could be as simple as 0.1s.
When using acceleration over time, all other components of F=ma must also be over time. F(t) = m(t) x a(t). Since mass change over the time period will be insignificant during 0.1s, we can use simple m.
We'll have F(t) = ma(t). This is substantially different than F = ma.
F(t) will be the average applied F over a time-period.
The torque, producing the F(t), will then be over a time-period.
I think this is where you need to alter your formula.
F = ma gives you the acceleration at an instant moment. Instant moment is when you get the F pulse from the engine. In other words when a cylinder fires.
To calculate the acceleration of the car you need acceleration over a time-period. It could be as simple as 0.1s.
When using acceleration over time, all other components of F=ma must also be over time. F(t) = m(t) x a(t). Since mass change over the time period will be insignificant during 0.1s, we can use simple m.
We'll have F(t) = ma(t). This is substantially different than F = ma.
F(t) will be the average applied F over a time-period.
The torque, producing the F(t), will then be over a time-period.
no. You can add as many other variables to both sides of the equation but if they still cancel out the result is the same.
F(x+y+z)=ma(x+y+z) is still the same as F=ma
F(x+y+z)=ma(x+y+z) is still the same as F=ma
Sorry but I don't agree, I'm with eclou here. F(t) = ma(t) is the same as F=ma. F(t) describes the instantaneous force at any time t. Now if we integrate this we get Force applied over some time. Since you are trying to separate the two here but they are not it reads as though you're trying to do F*t.
An engine producing 300 ft.lb @ 3000 RPM and 300 ft.lb @ 6000 RPM is substantially different.
Since during 0.1s, the 3000 rpm will generate an half the average torque than the 6000 rpm.
Therefore the 6000 RPM will accelerate twice as fast. The acceleration will be a result of generated torque and RPM, which is HP.
A 150ft.lb @ 6000 RPM will be the same as 300 lb.ft @ 3000 RPM. Both will generate the same torque at the wheels and both will generate the same HP at the engine.
However, if you don't alter the formula and stick to F = ma. It will tilt out the RPM out of the equations and would cause the acceleration being a pure result of Torque and not HP. This will describe the instant acceleration of each combustion.
Since during 0.1s, the 3000 rpm will generate an half the average torque than the 6000 rpm.
Therefore the 6000 RPM will accelerate twice as fast. The acceleration will be a result of generated torque and RPM, which is HP.
A 150ft.lb @ 6000 RPM will be the same as 300 lb.ft @ 3000 RPM. Both will generate the same torque at the wheels and both will generate the same HP at the engine.
However, if you don't alter the formula and stick to F = ma. It will tilt out the RPM out of the equations and would cause the acceleration being a pure result of Torque and not HP. This will describe the instant acceleration of each combustion.
I don't find this exactly true. During the power stroke the force is not instantaneous, but occurs over some period. Also, it will not appear as a square wave due to overlap in igntion cycles. It will have some oscillation yes, but it will be sinusoidal, not a step function.
03-23-2009, 07:41 PM
#424
Drifting
Work is defined as the change in energy; work and kinetic energy are measured using the exact same units, one is just the delta of the other. Hp-s is not typically one I'd use for energy, but it's just as valid as kWh or Joules or BTUs. When starting from 0mph (like in this example), Work and Ek are identical. I just didn't bother to write the delta and -1/2m0^2 part of the equation.
03-23-2009, 07:58 PM
#425
Rennlist Member
Thread Starter
ok, your right. had to think about that one for a sec! So, why didnt the numbers work out in the end?
My point was you were using HP x seconds and equating it to something that was going to be in J. It didnt look like it was in the right units. Now that i look, you did do KW and were looking at speed in m/s.
Heck, I could always use a physics lesson! its been awhile, but some of the stuff has stuck in the grey matter above.
It looked like you found the error in the end anyway and we agree. same engine HP , same rear wheel torque through the gear ratios. I still would like to know how we make those kinetic energy numbers work!
mk
Sjfehr said:
Now, I used engine horsepower which I know is wrong; I didn't expect to have to literally half that HP to the wheels to get the plots to match, though! In the end, though, I think I proved myself wrong. I was expecting to see the actual acceleration as the near-arrow-straight line to around 120mph that I recall from my owners manual, while knowing that the actual Ek chart is parabolic. I'm enough of a scientist to admit when the data has disproved my hypothesis, though. You're right; the data matches a pure and simple kinetic energy plot. It also shows that the C4S was considerably slower to 30mph than the numbers would otherwise indicate, which is, I imagine, a direct result of poor gear ratio when starting from a stop.
Kibort Quote:
Wrong, dead flat wrong. 250ftlbs and 500ftlbs can produce the exact same acceleration if the RPMs allowed are in proportions. (i.e. 500ft-lbs at 4000rpm vs 250ftlbs at 8,000rpm). This is so easy to prove. Try and find one actual instance where this is not true.
Not when both engines are running the same gearing at 4000rpm they won't! Which was my point. Unfortunately, it doesn't really add anything to the debate since HP are drastically different between the two engines in this example, so I'll let this one lie.
My point was you were using HP x seconds and equating it to something that was going to be in J. It didnt look like it was in the right units. Now that i look, you did do KW and were looking at speed in m/s.
Heck, I could always use a physics lesson! its been awhile, but some of the stuff has stuck in the grey matter above.
It looked like you found the error in the end anyway and we agree. same engine HP , same rear wheel torque through the gear ratios. I still would like to know how we make those kinetic energy numbers work!
mk
Sjfehr said:
Now, I used engine horsepower which I know is wrong; I didn't expect to have to literally half that HP to the wheels to get the plots to match, though! In the end, though, I think I proved myself wrong. I was expecting to see the actual acceleration as the near-arrow-straight line to around 120mph that I recall from my owners manual, while knowing that the actual Ek chart is parabolic. I'm enough of a scientist to admit when the data has disproved my hypothesis, though. You're right; the data matches a pure and simple kinetic energy plot. It also shows that the C4S was considerably slower to 30mph than the numbers would otherwise indicate, which is, I imagine, a direct result of poor gear ratio when starting from a stop.
Kibort Quote:
Wrong, dead flat wrong. 250ftlbs and 500ftlbs can produce the exact same acceleration if the RPMs allowed are in proportions. (i.e. 500ft-lbs at 4000rpm vs 250ftlbs at 8,000rpm). This is so easy to prove. Try and find one actual instance where this is not true.
Not when both engines are running the same gearing at 4000rpm they won't! Which was my point. Unfortunately, it doesn't really add anything to the debate since HP are drastically different between the two engines in this example, so I'll let this one lie.
hahaha, NOW who needs a physics lesson?
Work is defined as the change in energy; work and kinetic energy are measured using the exact same units, one is just the delta of the other. Hp-s is not typically one I'd use for energy, but it's just as valid as kWh or Joules or BTUs. When starting from 0mph (like in this example), Work and Ek are identical. I just didn't bother to write the delta and -1/2m0^2 part of the equation.
Work is defined as the change in energy; work and kinetic energy are measured using the exact same units, one is just the delta of the other. Hp-s is not typically one I'd use for energy, but it's just as valid as kWh or Joules or BTUs. When starting from 0mph (like in this example), Work and Ek are identical. I just didn't bother to write the delta and -1/2m0^2 part of the equation.
03-23-2009, 08:26 PM
#426
Rennlist Member
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03-23-2009, 08:49 PM
#429
Rassel,
Sorry but I don't agree, I'm with eclou here. F(t) = ma(t) is the same as F=ma. F(t) describes the instantaneous force at any time t. Now if we integrate this we get Force applied over some time. Since you are trying to separate the two here but they are not it reads as though you're trying to do F*t.
Sorry but I don't agree, I'm with eclou here. F(t) = ma(t) is the same as F=ma. F(t) describes the instantaneous force at any time t. Now if we integrate this we get Force applied over some time. Since you are trying to separate the two here but they are not it reads as though you're trying to do F*t.
We need average acceleration over a specific time. <- Is it here we don't agree?
If not, time is not moving and we're not moving either. So:
Average Acceleration between t=0 and t=1
Since we take average acceleration over a time period, we have to use average force over the time period.
Average force applied between t=0 and t=1
Average Force = RPM x F x k (k= konstant, depending on amount of cylinders and combustion cycle form) <- Or is it here we don't agree?
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Now, we can actually throw out the gearbox from the discussion. F is what causes the acceleration, it comes from the cylinders and if we have no losses, there is no need to add more variables.
To do a practical example: A normal engine will have a Four-Stroke Otto-Cycle.
For every 4 rounds, it will fire once, creating the Force F. This travels through the piston creating the Torque T.
Lets say we have one cylinder engine rotating at 1000 RPM. That will during one minute create an average Force of 1000 x F / 4.
If we increase the RPM to 2000 RPM. It will create an average Force of 2000 x F/4.
So, over time of one minute of constant acceleration. The average Force over time created will be twice when increasing the RPM from 1k to 2k. Hence the acceleration of the vehicle should be twice.
As mentioned before. An engine is basically an air pump. Why wouldn't it make a difference if its expanding twice the amount of air?
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If we don't count the acceleration as average over time, but take the acceleration for an instant moment. It will be a result of one single combustion causing F.
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Technically, "the amount of instant accelerations" are not static over time. But a result of how many of them you make over time.
Now, if we don't agree on the above. We can't agree on the below. If the number of Otto Cycles over time are irrelevant and only the strength of them accounted for, the conclusion would be torque -> acceleration.
I don't disagree with this at all until we get to the last statement that with RPM out the results are different and this I do not agree with. A dyno will measure the torque output. This is the actual engine torque output of the motor; assuming an engine dyno measuring at the crank. At that point we have a large difference between the two motors you've described, but once we gear them down and get them to the same speed at the rear wheels it no longer matters. Assuming we've geared them to the exact same wheel speed we will have the exact same torque output, so we have removed any effect of the difference of RPM.
I don't find this exactly true. During the power stroke the force is not instantaneous, but occurs over some period. Also, it will not appear as a square wave due to overlap in igntion cycles. It will have some oscillation yes, but it will be sinusoidal, not a step function.
03-23-2009, 09:12 PM
#430
Drifting
03-23-2009, 09:44 PM
#431
Three Wheelin'
Rassel, I think for the most part every were talking about is essentially the same and that if this were discussion were in person this would go much quicker and easier, but this is the best we can do.
This here might the one viewpoint we don't quite share. I understand your point and agree with it, but to a point and this maybe due to a differing viewpoint. I'm looking it as the application inwhich we can create a simple vehicle model.
From the combustion standpoint I have no clue what the force is or what the exact input to the crank is involved and for the application don't care. What I do know is that I can measure the output of the crank. This curve then defines the force that I can later impart to the rear wheels. What I have then is the definition of torque as a function of time, T(t), and therfore I now have a function for the Force that can be applied at the rear wheels F(t). Since F(t)=ma(t) and we now have F(t) we can find the function for a(t). With this you can do a lot and easily find the acceleration over time like you had discussed.
Simplied, yes I do see torque -> acceleration, but this is because I have the torque output for all time, at any instant, and as described can directly relate this to the force and acceleration.
Now, if we don't agree on the above. We can't agree on the below. If the number of Otto Cycles over time are irrelevant and only the strength of them accounted for, the conclusion would be torque -> acceleration.
From the combustion standpoint I have no clue what the force is or what the exact input to the crank is involved and for the application don't care. What I do know is that I can measure the output of the crank. This curve then defines the force that I can later impart to the rear wheels. What I have then is the definition of torque as a function of time, T(t), and therfore I now have a function for the Force that can be applied at the rear wheels F(t). Since F(t)=ma(t) and we now have F(t) we can find the function for a(t). With this you can do a lot and easily find the acceleration over time like you had discussed.
Simplied, yes I do see torque -> acceleration, but this is because I have the torque output for all time, at any instant, and as described can directly relate this to the force and acceleration.
03-25-2009, 10:57 PM
#432
Rennlist Member
Who won the torque debate?
SundayDriver.
Professional Racing and Driving Coach
SundayDriver.
Professional Racing and Driving Coach
03-26-2009, 02:01 AM
#433
Rennlist Member
Thread Starter
Hardly! Newton always wins the torque debate!
HP determines torque at the rear wheels at any vehicle speed, ANYWHERE, ANYTIME!
Now, do high torque engines of the same HP have a broader HP curve? Usually, Yes. (To your Point VR) BUT, it is NOT, and is in no way, the rule.
Ive shown two graphs, both actual and one based on Dez's to prove it. Yet, you would elect to ingnor it.
Now, Sunday actually bordering on wrong, due to his "flat torque" being an advantage. Many peaky hp engines have flat torque curves. by definition, a flat torque curve means HP is rising, until it isnt rising anymore
Usually, a brute force high torque V10 will have a broad HP curve, which means the usuable torque curve is really on the downward slope. remember, at 5250rpm, torque has to equal hp, and go down from there!
mk
HP determines torque at the rear wheels at any vehicle speed, ANYWHERE, ANYTIME!
Now, do high torque engines of the same HP have a broader HP curve? Usually, Yes. (To your Point VR) BUT, it is NOT, and is in no way, the rule.
Ive shown two graphs, both actual and one based on Dez's to prove it. Yet, you would elect to ingnor it.
Now, Sunday actually bordering on wrong, due to his "flat torque" being an advantage. Many peaky hp engines have flat torque curves. by definition, a flat torque curve means HP is rising, until it isnt rising anymore
Usually, a brute force high torque V10 will have a broad HP curve, which means the usuable torque curve is really on the downward slope. remember, at 5250rpm, torque has to equal hp, and go down from there!
mk
03-26-2009, 09:58 AM
#435
Rennlist Member