What's the mathematical(or otherwise) relationship of horsepower to torque and vice verse? I've seen #'s greater in hp than torque and vice versa. How is that possible?

 

http://www.elec-toolbox.com/Formulas/Motor/mtrform.htm

Those are a bunch of calculations, but most people just use this one that calculates HP based on TQ:

HP = (TQ x RPM)/5252

I've had it proven to me before, but I can't think of it off the top of my head, but you can see why HP always equals TQ at 5252 RPM no matter what combination it's in.

 

That's what I'm now trying to figure out. 33,000 ft/lb of torque=hp. So, hp is just a larger quantity of torque, correct? I'd still like to know why hp equals torque at 5252 and torque then exceeds hp after 5252 regardless the engine.

 

This will explain everything

http://www.hardtail.com/techtips/hpandtorque.html

Torque is the amount of work being done. HP is how quickly the work is done.

 

Thanks Stephen. I'm kind of confused on gearing after reading that article though. Why do performance cars go to bigger gears in the rear end? I know it help with off the line(60 ft times). Does it allow you to shift gears later due to less rpm's needed to turn them? That would allow you to pull longer, correct?

 

They use the higher gears for a few reasons. You need them with taller tires. It gets them into they're operating RPM range faster. I think it would keep you in the gears for a shorter period of time, but correct me if I'm wrong.

 

Gears and tire height depend on the car's shift point. For drag racing, it's all set up so that you're at the shift point when you cross the finish line or just before it. The car is maxed right out and won't go any faster when it's crossing the finish line.

My car will only go 140 mph. There are street cars that can beat that on the highway but I'm at 140 mph in a 1/4 mile while a street car may take 3/8 to 1/2 mile to get there.

Deep gears are needed to help accelerate the mass of the vehicle quickly. A lighter vehicle doesn't need as deep a gear. The faster the weight can get moving, the quicker the run is. Going too deep can easily create enough torque to overpower the tires though.

Look at things like Top Fuel dragsters. They do a 1/4 mile at 330 mph. At the 1/8 mile mark, they're already doing 280 mph because they're accelerating so fast. Now an 8000hp engine is also producing so much torque and the vehicle doesn't weigh very much so they only have one gear ratio choice. By the rulebook, they all have 3.20 gears in the diff.

 

That part is the other way around. Horsepower will exceed torque after 5252.

 

Yes. Horsepower is always less than torque below 5252 RPMs and always higher than torque above that point.

 

Yes, that's the way I meant to ask it. Anyone?

 

Definition: 1 horsepower (HP) = 33,000 lb-ft per minute. HP is calculated from measured torque and rpm.

Here's the math to get the HP formula. Torque is a force times a distance, or lb-ft. So to convert 33,000 lb-ft per minute to rotation, use 2 PI radians = 1 revolution. (Circumference of a circle = 2 PI x radius, and a radian is the distince around the circumference of the radius). 2PI = 2 x 3.1415.. or 6.2832.... Now 33,000/6.2832 = 5252

That gives the formula: HP = (Torque x rpm) / 5252

If the terms are rearranged to solve for Torque:

Torque = (HP x 5252)/rpm

At 5252 rpm HP = Torque exactly. Above 5252, HP is greater than Torque, and below 5252 Torque is greater than HP just due to the math formula.

An example: An 850 HP @ 9000 rpm Nextel Cup engine has:

Torque = (850 x 5252) /9000 = 496 lb-ft (@9000)

If another engine produced 496 lb-ft @ 4500 rpm it would have:

(496 x 4500)/5252 = 425 HP

Any help?

 

What you're saying is that without torque there is no horsepower, because horsepower is a calculated number based on the torque/force that the engine applies to the crank?

So every engine is a compromise based on the requirements of the vehicle in question?

Denny

 

Yes. HP is only a function of Torque. Torque and RPM is the only thing measured by a dyno. HP is simply calculated.

And yes. The compromise is based on what you're doing with the vehicle. You wouldn't cam an engine for high RPM torque (like a nextel cup engine) for towing your racecar to the track, just like you wouldn't put that tow vehicle's cam into an engine for racing around a road course -- very simple example. The main compromises are streetability and power range (rpm).

 

False. We've had these discussions here before, the search function is a wonderful thing.

 

Well, a brake dyno (which most engine dynos are) measures brake torque and rpm and power is calculated. On the other hand, inertia dynos, either chassis or engine, effectively measure the change in rpm of the inertia drum over very short periods of time, and calculate the power it takes to do that. Torque can then be backed out mathematically.

It gets complex when a brake (eddy current or water) is added to an inertia dyno, but that's another discussion. Lots of shades of gray here.

FWIW, when most folks talk about torque or power of an engine they are referring to brake torque or brake power, which means it is actually under load, not just free reving under no load. Of course the engine is producing torque/power internally if it is running. This is called Indicated torque or power. If there is no load on the engine, and the engine is "free reving" at a constant rpm, the Indicated torque/power is just enough to overcome the friction torque/power and the pumping losses.

One way to get more brake torque/power from an engine without increasing indicated power (which is determined by how well the engine pumps air and burns the air/fuel) would be to reduce friction losses and pumping losses. That sounds obvious, of course, but it becomes Job 1 in some high end competition engines. Cup restricter plate engines might be the ultimate example of this.

My $.02