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Old 03-04-2016, 11:00 PM   #81
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What?
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Old 03-04-2016, 11:01 PM   #82
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Originally Posted by 18psi View Post
if it was only about torque and nothing else wouldn't all dragsters be diesel powered?

Because it only about torque at the tyres, which is a factor of horsepower and gearing.

why do top fuel drag cars rev so high?

To make power. See above.

why not just make 3,000tq at 3k like a diesel and gear it accordingly?

See above.
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Old 03-04-2016, 11:02 PM   #83
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Originally Posted by fooger03 View Post
Not sure if retard or troll.
NASA champion and automotive engineer, IIRC. But please, call him a retard/troll some more. It makes your argument that much less believable.

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On the first half of this point, we can agree, on the second half of this point, we can disagree.

Engine torque is far from irrelevant. Acceleration depends entirely on wheel torque. Wheel torque depends on power (which is engine torque and engine speed) plus gearing.
What Jason (and I) are trying to say is that engine torque and acceleration have no correlation.. You can double or triple your engine torque, but if you only make that torque at 1/2 or 1/3 the engine speed, the power output remains the same, and acceleration will be the same. So yes, [i]engine[i] torque, as measured at the flywheel, is irrelevant.

If you change engine torque while maintaining the same power, acceleration will not change. If you change engine power while maintaining the same engine torque, acceleration will change. Thus, engine torque and acceleration have no correlation.

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Acceleration is independent of vehicle speed and dependent on wheel torque.
Sure, but only because wheel torque and engine power rise and fall in lock-step. You cannot increase wheel torque without increasing engine power at a given road speed.

Quote:
Acceleration will indeed dramatically reduce with engine speed *if power is constant*, because in order to keep power constant while engine speed increases we have to dramatically reduce engine torque.
This is completely false.

If engine power is constant, and the car is operating in a magical universe where there is no such thing as rolling resistance, drivetrain losses, or aerodynamic drag, then the car will accelerate at a constant rate.

The definition of power is a unit of energy consumed per time. Applying 200hp of energy to a body will cause that body to convert that energy into momentum. There is no engine speed at which the conversion of this energy into momentum will become less efficient, or happen at a slower rate, or cease to happen at all.

If that paragraph confused you, think about it this way. Let's say you have a hypothetical car, and it's operating in the hypothetical universe we're talking about (no aero drag, no rolling resistance, no drivetrain losses, perfect efficiency). It is making a flat 10ft.lbs of torque at the flywheel, regardless of engine speed. (It's a very small engine, able to rev infinitely high.) At 5252rpm, you have 10hp available to you. At 52,520rpm, you have 100hp available to you. Will the car not accelerate harder with 100hp than it will with 10hp?

Multiply RPM by 10 again. 525,250rpm, 1,000hp. Still just 10ft.lbs of engine torque, and yet the car is now making a thousand horsepower. Can you honestly say that the car will not accelerate at a higher rate with a thousand horsepower than it will with just ten horsepower? Remember, engine torque has not changed at all.

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We can also say that generally as engine speed increases, acceleration dramatically reduces *relative to power*
This is also completely false, for all the reasons explained above.
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Old 03-04-2016, 11:09 PM   #84
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Quote:
Originally Posted by nitrodann View Post
.
see, I was trying to get a few things clarified because a lot of things are assumed
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Old 03-04-2016, 11:11 PM   #85
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Originally Posted by speedengineer View Post
Consider:
P=F*v (power = force*velocity)

...

2) Acceleration is not constant with vehicle speed even with constant power. It dramatically reduces with speed, due to the 1/v component of the equation. This is a hyperbolic function.
Isn't the velocity in the above equation referring to engine speed, not ground speed?

Power = Horsepower
Force = Torque
Velocity = RPM

Thus, the equation (with the 5252 correction factor because units) becomes:

Horsepower = Torque * RPM
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Old 03-04-2016, 11:17 PM   #86
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Quote:
Originally Posted by Savington View Post


What Jason (and I) are trying to say is that engine torque and acceleration have no correlation.. You can double or triple your engine torque, but if you only make that torque at 1/2 or 1/3 the engine speed, the power output remains the same, and acceleration will be the same. So yes, [i]engine[i] torque, as measured at the flywheel, is irrelevant.

Acceleration will increase if nothing else is changed, however you will be going slower.

If you change engine torque while maintaining the same power, acceleration will not change.

Again I disagree, see above. Your statements rely on the gearing changing by the exact ratio that RPM/torque is being changed in your scenario. If the ratios are constant and so is output power over rpm, increases in engine torque implies lower rpm, which implies lower road speed, however wheel torque will increase as will acceleration.

If you change engine power while maintaining the same engine torque, acceleration will change. Thus, engine torque and acceleration have no correlation.

If you do this, you are changing the velocity of the car only, of course wind resistance will affect results.

Sure, but only because wheel torque and engine power rise and fall in lock-step. You cannot increase wheel torque without increasing engine power at a given road speed.



This is completely false.

If engine power is constant, and the car is operating in a magical universe where there is no such thing as rolling resistance, drivetrain losses, or aerodynamic drag, then the car will accelerate at a constant rate.

Despite the rotational force (torque at the wheels) decreasing as rpm rises?

The definition of power is a unit of energy consumed per time. Applying 200hp of energy to a body will cause that body to convert that energy into momentum. There is no engine speed at which the conversion of this energy into momentum will become less efficient, or happen at a slower rate, or cease to happen at all.

This statements here, this is where the confusion exists.

If that paragraph confused you, think about it this way. Let's say you have a hypothetical car, and it's operating in the hypothetical universe we're talking about (no aero drag, no rolling resistance, no drivetrain losses, perfect efficiency). It is making a flat 10ft.lbs of torque at the flywheel, regardless of engine speed. (It's a very small engine, able to rev infinitely high.) At 5252rpm, you have 10hp available to you. At 52,520rpm, you have 100hp available to you. Will the car not accelerate harder with 100hp than it will with 10hp?

No it wont because the force at the tyre where it meets the road has not changed one iota.

Multiply RPM by 10 again. 525,250rpm, 1,000hp. Still just 10ft.lbs of engine torque, and yet the car is now making a thousand horsepower. Can you honestly say that the car will not accelerate at a higher rate with a thousand horsepower than it will with just ten horsepower? Remember, engine torque has not changed at all.

And because gearing hasnt changed either has the force on the tyres. still only 10 ft lb pushing against the tarmac.



This is also completely false, for all the reasons explained above.
???

Last edited by nitrodann; 03-04-2016 at 11:27 PM.
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Old 03-04-2016, 11:23 PM   #87
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If power is constant, and power is work/time.

Then as we increase RPM, the car travels further per unit of time.

If power is flat from 5000rpm>10,000rpm then torque is halved at 10,000rpm.

At 10,000rpm we have the same amount of work achieved per unit of time compared to 5,000rpm because horsepower is the same, but that same amount of work is stretched in half because it has to move the car double the distance.

Work achieved is how far the car is moved.

If the same amount of power has to move the car twice as far in the same unit of time, how can it possibly have enough energy left over to accelerate it just as forcefully as at 5,000 rpm when that same amount of power only had to do half of the work in a given period of time?
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Old 03-04-2016, 11:28 PM   #88
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Hows this for an even easier example.

If I put smaller rolling diameter tyres on the car does it accelerate with more Gs or less G's in the same gear and at the same rpm and why?
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Old 03-04-2016, 11:29 PM   #89
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Quote:
Originally Posted by nitrodann View Post
Because torque is dropping.

The premise of your assertion above is what this whole thread is here to disprove.

I know that. I was wanting Fooger to answer that question with any other possible explanation. I was trying to make a point.
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Old 03-04-2016, 11:32 PM   #90
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Lol, sorry, I didnt realize.
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Old 03-04-2016, 11:53 PM   #91
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Hypothesize hypothesize hypothesize. Well on paper it's this or that. Blah blah blah.

Real world example with first hand knowledge of vehicles I've owned.

600hp big block v8 Chevy in a ~5000lb truck

600hp diesel I6 Cummins in a ~7200lb truck

Truck with the Cummins accelerated SUBSTANTIALLY harder than the v8 truck. Basically the same gearing.

Guess which one had more torque and less rpm?
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Old 03-05-2016, 12:02 AM   #92
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You were accelerating harder in the high torque car but at lower road speed for a given gear ratio. One can only assume.
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Old 03-05-2016, 12:52 AM   #93
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Quote:
Originally Posted by Savington View Post
Isn't the velocity in the above equation referring to engine speed, not ground speed?

Power = Horsepower
Force = Torque
Velocity = RPM

Thus, the equation (with the 5252 correction factor because units) becomes:

Horsepower = Torque * RPM
Nope, it actually is road speed! It's really cool how the math works out. A good analog would be T=I*alpha and F=m*a. Or P=T*rpm and P=F*V. Or you could look at it from the standpoint that Energy = force*distance. Take the time derivative and you have Power (time rate of energy) equals force times velocity (time rate of distance). P=F*v

The second part of the equation that relates acceleration to power, mass, and road velocity is equally interesting. It shows that acceleration naturally drops as vehicle speed increases (1/velocity). Acceleration drops in real life not only due to aerodynamic drag and rolling losses, but also just due to the physics! Another way to think of is to consider the kinetic energy of a vehicle, which is 1/2*mass*velocity^2. At higher velocities, more energy is required for a 1mph increase in speed. Thus for a given power, more time is required for that larger increase in energy, thus acceleration must be lower.
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Old 03-05-2016, 12:55 AM   #94
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Quote:
Originally Posted by nitrodann View Post
You were accelerating harder in the high torque car but at lower road speed for a given gear ratio. One can only assume.
You can assume anything you want.

Dodge had a 48re which is geared 2.45, 1.45, 1 and 35" tires with a 3.73 axle ratio.

Chevy had a turbo 400 which is geared 2.48, 1.48, 1 and had 35" tires with a 3.73 axle ratio.

From a stop, how exactly would one accelerate harder but have a lower road speed?

Trust me, the dodge had more road speed at any given time interval during the first 12 seconds. It also had more than double the torque with the same basic hp.
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Old 03-05-2016, 12:56 AM   #95
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because the higher speed spreads the power applied out over a greater distance right?
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Old 03-05-2016, 01:06 AM   #96
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Quote:
Originally Posted by nitrodann View Post
because the higher speed spreads the power applied out over a greater distance right?
No again.

Dodge - 110mph give or take in 12 seconds covering a distance of 1320ft

Chevy - 95mph give or take in 14 seconds covering a distance of 1320ft

Keep in mind there is also a 2000lb weight difference.
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Old 03-05-2016, 01:07 AM   #97
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why is it then?
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Old 03-05-2016, 01:11 AM   #98
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Quote:
Originally Posted by nitrodann View Post
why is it then?
600ftlb vs 1480ftlb.
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Old 03-05-2016, 01:20 AM   #99
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because the higher speed spreads the power applied out over a greater distance right?
No again.


This question, why no
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Old 03-05-2016, 01:51 AM   #100
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Gutted, your truck example is utterly flawed. Using the same gearing for two engines so drastically different is not appropriate. Had the chevy been geared such that it was making use of it's higher RPM it would have in fact accelerated faster than the heavier Dodge.

I think it helps if you use a CVT in your hypothetical car so the cars stay right on their power peaks, and you can avoid worrying about area under the curve which requires calculus.

Two cars, same weight, aero, everything, engines have the same peak power, but one does it at twice the RPM and therefore half the torque of the other, using a CVT will accelerate the same. Yes as the speed increases the acceleration will fall off, but it will fall off the same for both cars, and it's shown in the equation from above where acceleration = power/(velocity * mass).

Or you could say the two cars are the same, except both make the same torque, but one makes it at twice the RPM and therefore twice the power, and again we use the CVTs keeping things at their peaks, the higher power car will be accelerating twice as hard.
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