Quote:
|
Originally Posted by SternoBoy
I thought some more about what you wrote and modified my thought experiment to this: Imagine you have one bitchin' CVT. Would it be better to hold the engine at a constant 100ft/lbs of torque at 8000rpm or a contant 95ft/lbs at 9000rpm? And the answer of course is 95ft/lbs at 9000 rpm because for any given final drive speed, while losing 5% torque at the engine, one would gain 12% through gear reduction ( 95ftlb * 9 / 8 = 106 ft/lbs).
So I take it back. You're right. Sorry I doubted you.
|
After reading what your post USED to say before you edited it (which is below), I performed the very same 'thought experiment' as you did
Excellent post, btw... it definitely helped me finalize my understanding of the relationship of Horsepower and Torque. So, for learning/understanding purposes, here's your pre-edited post and how I think both of us came to the same conclusion:
Quote:
|
Originally Posted by SternoBoy
Follow the math with me:
In my earlier examples of peak horsepower of at 163 @ 9000rpm (and 95 ft/lbs tq), a 9:1 total drive ratio resulted in a thrust of 789lbs and an acceleration of 0.39G, but with the same 9:1 total drive ratio and the peak torque of 100 ft/lbs at 8000rpm, the resulting thrust would be 830lbs accelerating at 0.44G (albeit at a lower final drive speed of 889 rpm). Thus, peak torque is where maximum acceleration occurs. (Again, in this thought experiment the thrust and acceleration figures are based on a 26" diameter tire and a total vehicle weight of 2000 lbs.)
Or more fundamentally, since a = f / m and torque is force, then maximum acceleration is going to happen at maximum force (torque) for any given gear ratio.
- Mike
|
As it turns out, your above calculation proves that you are correct that:
Maximum Acceleration occurs at Maximum ENGINE Torque... IF GEARING IS HELD CONSTANT.
However, this results in the two numbers being compared AT DIFFERENT SPEEDS because, with gearing held constant, the car running at a greater RPM will be going at a proportionally higher speed.
So, I solved for the gear ratio that would make the car @ 9000rpm run at the same speed as the car @ 8000rpm (which had a final drive speed of 889rpm).
9000 rpm / 889 rpm = 10.1237:1 ratio
So, solving for the thrust, I used the thrust calculated by the old final drive ratio (9:1), divided by the old final drive ratio (9:1), and multiplied by the new final drive ratio (10.1237:1).
789 lbs thrust
(@ 9:1) / 9
(old FD ratio) * 10.123
(new FD ratio) = 887.514 lbs thrust
SO, given a CONSTANT SPEED (and OPTIMIZED GEARING), the car with more HORSEPOWER, not Torque, produces the most Thrust (Applied Torque), becasue it can multiply the torque by a greater gear ratio.
So, just for fun, I decided to compare the Thrust and Horsepower ratios for the car @ 9000rpm and the car @ 8000rpm:
887.5 lb thrust
(@ 9000 rpm) / 830 lb thrust
(@ 8000 rpm) =
1.069
HP @ 100tq/8000rpm = 152.3
162.75 hp
(@ 9000 rpm) / 152.3 hp
(@ 8000 rpm) =
1.069
Therefore, because the ratios of thrust and hp between the two cars are the same, we can conclude that:
HORSEPOWER is directly proportional to an engine's POTENTIAL to produce THRUST.
Therefore, it follows that,
At any given SPEED, assuming OPTIMIZED GEARING, the car producing more HORSEPOWER will produce more THRUST proportional to it's horsepower advantage.
_________________
Well, I think we are finally in agreement here, SternoBoy... And I think real answer was somewhere between what each of us originally thought.
I've still got a lot fine-tuning to do with my thoughts, and I've got to correct any previous statements to align with my new understanding... and hopefully eventually be able to provide a simple explanation, because I KNOW that the above is not going to cut it