HP vs Torque vs Fuel Efficiency

[ksa8907]

wow, lots to read. maybe im wrong, but i don't care much about hp ratings and numbers, torque is a more accurate description imo. and gearing? it's just torque manipulation. the engine goes a farther distance (revolutions) to gain a mechanical advantage.

[mort]

Originally Posted by niky

If you use a rocket instead of a drive wheel and drive your bike in space, you will be able to maintain perfect acceleration no matter how much momentum you put into the bike.

This is nonsense. It takes power to accelerate in space just like on earth. If your engine produces 100 hp and your rocket mass is 600 lb your acceleration at 100 mph will be .625 g. At 1000 mph acceleration will be .0625 g

-mort

[sendler]

I haven't seen an acceleration chart like that before though. Thanks for posting it. It's interesting to see how high the value in 1st peaks peaks and drops back down before you reach the shift point to second. And what a loss of acceleration it is as you are forced to chose each higher gear. Even at what is eventually the same rpm range.

[darcane]

Originally Posted by mort

Hi sendler,
No, it is the road speed that matters the most. Power makes the thing go and the amount of power required at any speed depends on the speed.
If you use the ecomodder calculator here (note: the aero and rolling resistance forces are shown in Newtons) and think motorcycle... you get 30 hp needed to keep going 100 mph.

But the amount of power it takes to accelerate also depends on your speed. The acceleration is just like any other force being applied to the bike. Let's say you want to accelerate at 1 g, and you are going 50 mph. Again according to the ecomodder calculator, that would require 80 hp.
But how much power does it take to accelerate at 1 g at 100 mph. Your speed has doubled so the power has doubled: 160 hp.

So comparing the acceleration at 10 mph and 60 mph means comparing one condition with another that will require 6 times as much power - put another way: The same amount of power available at 10 mph and 60 mph will deliver 1/6th the acceleration.


Also I think the ecomodder calculator is wrong.
-mort

This is incorrect. Your version of physics would make setting land speed records trivially easy.

The power required to overcome aerodynamic drag is not linear, but is instead a cubic function of speed. If it takes 20hp to overcome aerodynamic drag at 50mph, it will take you 160hp to overcome drag at 100mph. If you double your speed, you need eight times (2^3) the power to overcome aerodynamic drag. You also have to factor in rolling resistance and friction, although those are much smaller factors at higher speeds.

So, if you accelerate at 1g at 50mph and it takes 80hp (20hp due to aero drag, 60hp you accelerate your mass at 1g, ignoring rolling resistance), at 100mph you should require 220hp (160hp due to aero drag, 60hp to accelerate your mass at 1g).

Double this hypothetical car's speed again and it will take 1340hp to accelerate at 1g at 200mph.

The ecomodder calculator is valid, but can only give you accurate results if you provide accurate inputs (which are not always easy to determine).

[sendler]

The examples with the pick up trucks is a bit of a trick question. If you knew the gear ratios (as he did on the CBR250R forum) and multiplied through from the engine torque you would see that at that moment the rear wheel torque would be the same in all three gears even though the gears are different and the rpms are different.
.
And from the acceleration chart I can see that I got sidetracked and that acceleration will be increased with increasing rear wheel torque even though that rear wheel torque will average higher when shifting either side of the power peak of the engine. Not the torque peak of the engine.

[darcane]

Originally Posted by sendler

The examples with the pick up trucks is a bit of a trick question. If you knew the gear ratios (as he did on the CBR250R forum) and multiplied through from the engine torque you would see that at that moment the rear wheel torque would be the same in all three gears even though the gears are different and the rpms are different.
.
And from the acceleration chart I can see that I got sidetracked and that acceleration will be increased with increasing rear wheel torque even though that rear wheel torque will average higher when shifting either side of the power peak of the engine. Not the torque peak of the engine.

I hadn't gotten to that question, but yes, if the horsepower is "flat", then acceleration will be the same regardless of which gear you are in.

That said, it is quite unusual to have a flat horsepower curve for a large enough rpm range to span multiple gears at the same speed.

[mort]

Originally Posted by darcane

This is incorrect. Your version of physics would make setting land speed records trivially easy.
...
So, if you accelerate at 1g at 50mph and it takes 80hp (20hp due to aero drag, 60hp you accelerate your mass at 1g, ignoring rolling resistance), at 100mph you should require 220hp (160hp due to aero drag, 60hp to accelerate your mass at 1g).

Hi darcane,
My physics is fine. You misunderstood the point of my post.
Yes, the power going to aerodynamic drag grows as the cube of the speed. But what you wrote (above) is wrong regarding acceleration.
Power is force multiplied by speed.
So if it takes 60 hp at 50 mph to accelerate at 1 g, then at 100 mph the power needed to accelerate at 1 g is 120 hp.

My point was that the reason you feel less oomph at high speed than at low speed is that if you only have the same amount of power - that power buys less acceleration the faster you go.

If you are going really fast, then aerodynamic drag uses up all your power, and you can't accelerate. But at relatively slow speeds, where the power to overcome rolling resistance and aero drag are low, then most of your power is available to accelerate. Consider if you could apply full power at 5 mph compared to 25 mph. If the power at the tires is the same then your acceleration at 25 mph will be 1/5th what it was at 5 mph. This why it is easy to burn rubber from a stop and impossible at 40 mph.

The actual physics: P = V * M * A
-mort

[sendler]

Originally Posted by mort

It takes power to accelerate in space just like on earth. If your engine produces 100 hp and your rocket mass is 600 lb your acceleration at 100 mph will be .625 g. At 1000 mph acceleration will be .0625 g

-mort

Good example. Showing a much greater difference in speed than I used in my example makes it very easy to see that given the same power and mass, the higher the speed, the less the acceleration. Before any frictional losses.

[darcane]

Originally Posted by mort

Hi darcane,
My physics is fine. You misunderstood the point of my post.
Yes, the power going to aerodynamic drag grows as the cube of the speed. But what you wrote (above) is wrong regarding acceleration.
Power is force multiplied by speed.
So if it takes 60 hp at 50 mph to accelerate at 1 g, then at 100 mph the power needed to accelerate at 1 g is 120 hp.

My point was that the reason you feel less oomph at high speed than at low speed is that if you only have the same amount of power - that power buys less acceleration the faster you go.

If you are going really fast, then aerodynamic drag uses up all your power, and you can't accelerate. But at relatively slow speeds, where the power to overcome rolling resistance and aero drag are low, then most of your power is available to accelerate. Consider if you could apply full power at 5 mph compared to 25 mph. If the power at the tires is the same then your acceleration at 25 mph will be 1/5th what it was at 5 mph. This why it is easy to burn rubber from a stop and impossible at 40 mph.

The actual physics: P = V * M * A
-mort

Bugger, I believe you are mostly right on this. I had a free-body diagram in my head to determine what forces would be acting on an accelerating car to account for everything. While the force would be the same in either case to get 1g acceleration (after subtracting the effects of aero and friction) the power would not, but would instead be proportional to velocity as you state.

However, your example is still misleading because the effects of aerodynamic drag increase exponentially and very quickly dwarf the power needed for acceleration.

[darcane]

Originally Posted by mort

This is nonsense. It takes power to accelerate in space just like on earth. If your engine produces 100 hp and your rocket mass is 600 lb your acceleration at 100 mph will be .625 g. At 1000 mph acceleration will be .0625 g

-mort

Nonsense?

Jet engines are rated in Thrust (which is really just a Force) rather than Power. With a constant thrust, power increases proportionally to speed. If you eliminate aerodynamic drag by being in space, then Acceleration is constant if Thrust is constant.

This is why he specified a rocket engine (which is a type of jet engine).