Note no RPM information above.

What is [Power = Work/Time]? Isn't that horespower? What is Time? Isn't that rpm?

Acceleration = Work / (Mass * Distance)

Acceleration = Power / (Mass * Distance * Time)

Hmmm, you're confusing me with this comparo of equations.


Bret, you sure you want to take this position? Alot of stored energy in flywheels.

 

Lets say I have a HP peak at 5500 rpm and my torque peak is at 3800rpm.... Where do I shift and why?

If I shift at 4400rpm I will move so slow that I will open the doors and start picking up coins from the road.

I see it now and its all in the gears... Its very probable that the torque multiplication in first gear at a bit past peak HP is still a lot more than at peak torque in the next gear. That is why you want to move to the HP band too...

Tell me Bret... Everything is a combo... a bad comboed engine will run like crap right? So its a combo of transmission and differential and rpm range for torque multiplication to the road...

Remember another thing... higher numerical gears will eat up more HP, but it takes a lot less time and speed to get to their peak HP and therefore accelerate quicker.

 

Power = Work/Time is just a way of expressing power. The rate at which work is done is power, just as the rate at which velocity changes is acceleration, etc.

Time would be measured in seconds like normal. RPM would be the rotational analogue of velocity (revolution, degrees, radians, etc. being the rotational equivalent of position)




Good catch, that was my bad - work = Power * Time (re-arrangement of the initial power = work/time equation), so that should read

Acceleration = Work / (Mass * Distance)

Work = Power * Time

Acceleration = (Power * Time) / (Mass * Distance)


But regardless the point that you can solve for acceleration without knowing RPM is borne out.




Exactly! and if you look at it from a "delivered" or multiplied torque setup you should shift when your 1st gear shift point multiplied torque = your 2nd gear multiplied torque.

This will be *exactly* the same point as where the hp values are equal, thus maximizing area under the curve.

Gearing, transmission, etc. are all important to the combination - that's why I kept saying "potential to accelerate" - with a CVT we could come pretty damn close to the ideal though.


Chris

 

Don't you need RPM to get the Power in the 3rd equation?

I think when this is over I will keep this thread in Word format and find a way to assimiliate it all into my wee brain hehe.

 

Power would just be horsepower. I can say 500 average horsepower and that is all you need to know.

Were I to give you a torque parameter you would have to insert it as rpm also (power = torque / RPM * K) - then you would have RPM information in there.

And remember, though rpm is a position over time (velocity) measurement, it is in the rotational frame of the motor - so a linear velocity, etc. would not be the same as giving RPM values.


Note that the formula I gave you above is not necessarily that useful - there are slightly more involved derivations that will give you the final values much easier (in a more useful manner) and include proportionality constants - such as

ET = ((Weight / HP)^(1/3)) * 5.825

MPH = ((HP / Weight)^(1/3)) * 234


These are still simplistic as the model they are based on doesn't take into account frictional losses, other deviations from non-ideal, wind resistance, etc. But they do further bear out the fact that only HP matters in this context.

But they should make it more clear that you need only hp (no rpm information) to determine *potential* to accelerate.


Chris

 

Call me kooky but what is the "*" in your equations represent (I'm hard headed).

 

the * is a multiplication symbol.

Chris, yeah it was late and I ment TQ. As you said TQ and RPM are what counts.

We need to look at more real life examples say:

two engines, same car.

2500lbs, quick change rear end, same size tires

800hp 8500rpm peak Winston Cup Motor

vs.

800hp 17000rpm peak F1 Motor.

both have the same flat torque cuve.

Or more exactly

494.30 ft lbs @ 8000rpm 358 cube small block

vs

247.15 ft lbs @ 17000rpm 3.0L F1 engine

one car runs a 4.10 gear and the other runs a 8.20 gear.

Which one is faster? You have much more gear in the F1 car,

Seems that if both have the same shape and relative value TQ cruve then they will be equal because they are both putting out 2026.63 ft lbs after mulitplication.

Highlander,

"Tell me Bret... Everything is a combo... a bad comboed engine will run like crap right? So its a combo of transmission and differential and rpm range for torque multiplication to the road..."

Yep there will be one combo that will be the quickest. The more gears the narrower the power band you have to work in the easier it is to do.

Arnie,

yeah, stored inertia in flywheels is pretty important, the rest of the system can be as light as possible. Low inertia helps HP and acceleration, but high inertia in a flywheel is beneficial on launch. (at least in a heavy Camaro, might not be the best thing in a Pro Stock car.)

Bret

 

Assuming that they will see the same average power down the 1/4 mile track (e.g. no wierd powerband issues, etc.) then both your cars should run the same. (Same hp / same delivered torque - however you want to look at it.)


Chris

 

Thanks for the clarification on the symbol. I need all the math help I can get .....lol

 

There is some really great info on this post but I think that alot of you are still missing the big picture. By the way this is the first time that I have heard of anyone on these message boards besides myself that realized the signifigance of drivetrain inefficiency and inertial losses that come from running numerically higher gears. I have had this debate several times and people seem to forget that the engine had to do work to accelerate its own mass and that friction goes up with the square of the speed increase. The big picture that I was referring to was that some people still seem to think that torque is what matters on a street car and that "hp sells engines". I think we can just lay it out and say that Horsepower is what wins races. You can have all the torque in the world and if you arent applying it at any appreciable rate then you wont be making any power. I agree that gearing is merely a method of maximizing the engines ability to use its power band to accelerate your car as fast as it can within its range of operation. A vehicle will accelerate proportional to its torque curve and even though its acceleration forces are greatest at torque peak this is not where it is applying the maximum power and in the end the power to the wheels is what is truly moving the vehicle. So I hope we can agree that power is all that matters and that you should focus on building your engine to produce the kind of power you want at the kind of rpm that will best suit your application.