Acceleration curve vs torque curve vs HP curve.
#63
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In theory, yes. In practice, I've spent lots and lots of time comparing RPM-calculated "speed" to GPS-verified speed, and they never, ever match. At high speed, the difference is substantial (5%+). RPM-calculated speed always overestimates actual ground speed. My hypothesis is deformation of the tire at high speed.
Tires increase in diameter at higher speed (centrifugal force - see Topfuel Dragster tires). So the speed calculated based on RPM should be underestimating at high speed, not overestimating, assuming that it was close at lower speed.
#64
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I am perfectly aware that I am going to disagree with the MT.net gods here. But:
Tires increase in diameter at higher speed (centrifugal force - see Topfuel Dragster tires). So the speed calculated based on RPM should be underestimating at high speed, not overestimating, assuming that it was close at lower speed.
Tires increase in diameter at higher speed (centrifugal force - see Topfuel Dragster tires). So the speed calculated based on RPM should be underestimating at high speed, not overestimating, assuming that it was close at lower speed.
#65
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Savington, what were you using for tire circumference in that calculation?
I ask because, the effective circumference is a fair bit less that what you get if you just calculate using the nominal tire dimensions.
Example:
I just picked a 224/45-15 Toyo R888 as an example, using the nominal dimensions I get a diameter of 22.97 inches
Per tire rack toyo claims the tire is 22.8 in diameter, so slightly shorter than it should be.
Using the 22.8 diameter calculating the revs per mile, I get 885 revolutions per mile, but again per tire rack the revs per mile is 911, indicating about 2.5% shorter yet.
I think this makes perfect sense because the tire is always squished on the bottom a little bit, I'm guessing by almost 2.5%
If I screwed up the math I'm going to be embarrassed.
I ask because, the effective circumference is a fair bit less that what you get if you just calculate using the nominal tire dimensions.
Example:
I just picked a 224/45-15 Toyo R888 as an example, using the nominal dimensions I get a diameter of 22.97 inches
Per tire rack toyo claims the tire is 22.8 in diameter, so slightly shorter than it should be.
Using the 22.8 diameter calculating the revs per mile, I get 885 revolutions per mile, but again per tire rack the revs per mile is 911, indicating about 2.5% shorter yet.
I think this makes perfect sense because the tire is always squished on the bottom a little bit, I'm guessing by almost 2.5%
If I screwed up the math I'm going to be embarrassed.
#66
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Tire treads flex, and rubber deforms - at higher speeds under no acceleration wind resistance is present in not-insignificant quantities - tires are still doing work. At the base of the equation then, the aerodynamic drag of the vehicle is pushing rearward on the drive tires as they spin forward. Since tires are not a rigid solid, the treads flex and compress while in contact with the ground, and then spring backwards and expand when contact ends. Since ground speed is lost due to the initial flex/compression at contact, but no speed is gained with the ground contact ends, it makes sense that under light loads, the tire must spin faster than the external circumference of the tire would indicate. The effect would be magnified under acceleration or with more flexible treads. The circumference of the tire should not see an increase of any significance if we are using steel belted radials since steel is so wonderfully inelastic under tension.
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Savington, what were you using for tire circumference in that calculation?
I ask because, the effective circumference is a fair bit less that what you get if you just calculate using the nominal tire dimensions.
Example:
I just picked a 224/45-15 Toyo R888 as an example, using the nominal dimensions I get a diameter of 22.97 inches
Per tire rack toyo claims the tire is 22.8 in diameter, so slightly shorter than it should be.
Using the 22.8 diameter calculating the revs per mile, I get 885 revolutions per mile, but again per tire rack the revs per mile is 911, indicating about 2.5% shorter yet.
I think this makes perfect sense because the tire is always squished on the bottom a little bit, I'm guessing by almost 2.5%
If I screwed up the math I'm going to be embarrassed.
I ask because, the effective circumference is a fair bit less that what you get if you just calculate using the nominal tire dimensions.
Example:
I just picked a 224/45-15 Toyo R888 as an example, using the nominal dimensions I get a diameter of 22.97 inches
Per tire rack toyo claims the tire is 22.8 in diameter, so slightly shorter than it should be.
Using the 22.8 diameter calculating the revs per mile, I get 885 revolutions per mile, but again per tire rack the revs per mile is 911, indicating about 2.5% shorter yet.
I think this makes perfect sense because the tire is always squished on the bottom a little bit, I'm guessing by almost 2.5%
If I screwed up the math I'm going to be embarrassed.
I don't really have a good hypothesis beyond "it flexes a lot" when it comes to explaining that difference, but the data was really consistent.
Last edited by Savington; 03-04-2016 at 11:55 AM.
#68
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Kind of late to the party here. Looks like Andrew and others have it covered.
Here is another way to look at acceleration from a very simple first principles approach. Neglecting aero drag and rolling losses, etc.
Consider:
F=m*a (force = mass*acceleration)
P=F*v (power = force*velocity)
Combine and solve for acceleration:
a=P/(v*m) (acceleration = power/(velocity*mass) )
This shows two important things:
1) Acceleration depends on power, and that engine torque is completely irrelevant.
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.
Here is another way to look at acceleration from a very simple first principles approach. Neglecting aero drag and rolling losses, etc.
Consider:
F=m*a (force = mass*acceleration)
P=F*v (power = force*velocity)
Combine and solve for acceleration:
a=P/(v*m) (acceleration = power/(velocity*mass) )
This shows two important things:
1) Acceleration depends on power, and that engine torque is completely irrelevant.
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.
#69
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Stumbled on an interesting (and relevant) video today, and remembered seeing this thread a few days ago. Thought I'd share.
It mirrors a lot of what has been said in this thread, but with the benefit of more visuals.
https://www.youtube.com/watch?v=UIQj...ature=youtu.be
It mirrors a lot of what has been said in this thread, but with the benefit of more visuals.
https://www.youtube.com/watch?v=UIQj...ature=youtu.be
Last edited by nick470; 03-04-2016 at 11:18 AM.
#70
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Kind of late to the party here. Looks like Andrew and others have it covered.
Here is another way to look at acceleration from a very simple first principles approach. Neglecting aero drag and rolling losses, etc.
Consider:
F=m*a (force = mass*acceleration)
P=F*v (power = force*velocity)
Combine and solve for acceleration:
a=P/(v*m) (acceleration = power/(velocity*mass) )
This shows two important things:
1) Acceleration depends on power, and that engine torque is completely irrelevant.
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.
Here is another way to look at acceleration from a very simple first principles approach. Neglecting aero drag and rolling losses, etc.
Consider:
F=m*a (force = mass*acceleration)
P=F*v (power = force*velocity)
Combine and solve for acceleration:
a=P/(v*m) (acceleration = power/(velocity*mass) )
This shows two important things:
1) Acceleration depends on power, and that engine torque is completely irrelevant.
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.
How is "F" created on most cars?
#72
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Not sure if retard or troll.
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.
Acceleration is independent of vehicle speed and dependent on wheel torque. 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. Unfortunately for you, the *if power is constant* thing is irrelevant, because generally *torque* is the more constant line on a dyno plot which forces power to increase.
We can also say that generally as engine speed increases, acceleration dramatically reduces *relative to power*
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.
We can also say that generally as engine speed increases, acceleration dramatically reduces *relative to power*
#79
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Before this becomes 3 pages of semantics, let me ask start with this.
You were being sarcastic in your above post, and you dont legitimately think we should all run diesels with 3000ftlb and 50hp, right?
You were being sarcastic in your above post, and you dont legitimately think we should all run diesels with 3000ftlb and 50hp, right?