Race cars - Torque vs Hp - The Undiscovered Country (for many)

Zanick, the matter hasn’t been reduced to a philosophical debate in the least. Your supposition concerning horsepower and torque is quite simply backwards. Torque isn’t a factor of horsepower rather; horsepower is a derivative of torque. You must first produce torque if you are to “calculate” a “horsepower rating” for the engine.
>>>>>>>>>>>>>>>>>>while this is true, when looking at strictly engine torque (not the resultant force or torque at the driven tires, it might not be a direct indicaton of accelerative potential, unless you know engine speed. So, HP ratings and curves can be a better indication of acceleration potential. Yes, at the engine Hp is the rate that the force does work. Its all about rate of doing work, which is hp but its cause by the force at the rear tires, not the engine. (meaning the engines numerical torque value can be misleading as a comparitive factor)

Horsepower is an intangible quantity that cannot be measured directly, but more importantly, horsepower isn’t a “force”. Horsepower is simply a rating of how much work can be performed over time and distance. It is “calculated” per the torque measured since torque CAN be measured directly in ft lbs. For instance, when an engine is stated as having 200 HP, it is merely a rating of how much work can be performed at a given RPM per the engine’s available torque.

>>>>>> agreed, but isnt power also the rate of change of kinetic energy as well. so , the rate of change of speed of the drums can be calculated to show HP without knowing torque.
I know that the force is makeing the acceleration, but as far as what causes the force, isnt that caused by the energy source and its potential energy? (a 75amp/second 10v battery that produces 1hp-second or 750watt-seconds? Ill tie back into this below.)

The dyno measures torque by applying a known braking force to the dyno’s rotating drum. When the vehicle’s drive wheels accelerate the dyno’s drum under this braking force up to a given RPM in a given time, it reveals the torque being applied by the vehicle’s drive wheels, as it would require a given torque to have accomplished this rate of drum acceleration in the time recorded.
>>>>>>>>>>for a brake dyno this is true, but there are inertial dynos that are just drums weighing up to 3000lbs and are free spinning. they only measure the acceleration change of the drum and calculate everything from that.

Force produced in a Combustion Engine

Consider where the combustion engine actually produces its force. It is per the optimized A/F ratio ignited per each power-stroke, as the piston and its connecting rod are forced downward by the expanding hot gases in the cylinder with the other end of the connecting rod applying its force to the crankshaft in a manner that provides leverage to rotate the crankshaft and produce crankshaft torque. So, torque is the accelerative component, not horsepower.
>>>>>>>>>>>>>>but doesnt the rate of expanding gasse the cause for the force? without the energy, power/t available, you dont have any force? correct?

You need to get horsepower out of your head and acknowledge the actual source of the vehicle’s acceleration; that being torque. Horsepower is only a rating of how much work can potentially be performed at a given RPM per the available torque.
>>>>>>>>>>>Doesnt the equation Acceleration=power/momentium point to acceleration being determined by power, which creates the force? this is where i was suggesting it gets a little bit philosophical. :) I understand that the force creates the acceleration. a=F/m. I get that. But, if we are going to dump 1 oz of gas in an engine and get the energy out of it, we can get lots of rpm and a little torque or visa versa. in the end, the force we will generate at the driven wheels will be multiplied through a gear box so that we maximise power, not engine torque at the rear wheels and the resultant torque at the rear wheels will be proportional to the power. This torque can be made from low engine torque or high engine torque, based on the RPM range needed to create max power at any particular vehicle speed.



Horsepower is a “rating” of potential work that can be accomplished, but it is NOT an actual force

For instance, my workouts had me moving approx 30,000 pounds of free weights a distance of 3 feet in an hour’s time. This can be given a horsepower rating HOWEVER, that horsepower rating does not imply an actual force used. It merely specifies a total of 30,000 pounds was moved 3 feet over a span of one hour (a given weight moved a given distance over time).

>>>>>>>you did work during the hour. it can be rated in watt-seconds, KWh or Hp-seconds with are unit measures of work, not rates of doing work. rates of doing work wold be if you then lifted 550lbs in 1 second 1 foot. that would be using 1 hp. we could then calculate the acceleration caused by that force over that period of time.


In fact, some weight movement was several hundred pounds per rep while things like curls were more around the 95 to 115 pounds. Again, horsepower doesn’t specify a given force rather; it implies a given amount of work that can potentially be accomplished.
>>>>>>> but the rate of which the force does work, is HP. you bench press 550lbs ove 1 foot in 1 second, you needed 1hp to do it. If you use pulleys to do it slower and it takes 4 seconds to do this work, then the HP required was only 1/4 hp, but the force at the weight was still 550lbs being lifted over 1 ft.

Hopefully this will provide the insight you need to resolve your confusion associated with horsepower. Unfortunately, those who tend to adore horsepower and have long confused its meaning have a most difficult time relinquishing their love for its mighty-sounding namesake, when it’s actually “torque” that should be held in highest accord.
>>>>>>>>>>>>This is ironic as ive been fighting the torque folks for 10 years in the field of racing, and im a very small minority. There are countless discussions on the topic, all feble arguments becuase in the end, torque is a part of HP. HP can determine torque at the rear wheels, even though torque at the engine multiplied through the gears does this too.
I understand that the force is what does work. or torque in the rotational plane. If we look at constant variable gear boxes, where will engines operate to maximize acceleration, peak engine torque or peak HP???? right, peak engine HP! this will yeild the maximum amount of torque at the rear wheels at any vehicle speed, right?

I demonstrated the example below back in my mini-bike building days. It demonstrates the significance of an engine’s torque.

Firsthand Example:

I used a 1.5 HP lawnmower engine to power a mini-bike and I geared it to yield a top-speed of 35 MPH while on flat ground with my body weight of 140 pounds. However, while riding at 35 MPH on flat ground, I came to one of the steepest streets in my neighborhood. My mini-bike steadily lost speed and would not make it all the way up to the top of the steep hill. When I hit the hill, the engine was at its upper rev range therefore producing its 1.5 HP, so why could it achieve 35 MPH on flat ground, but not make it to the top of the steep hill?
>>>>>>>>>>>>>>>becuse the hp requried to climb that hill was not availble from the 1.5HP engine. HP is the rate of doing work. that amount of fuel and air was not enough to do the job of lifting you and the mini bike up the hill at that speed. Power=Forcex speed. so , you slowed, got out of the power band and could have come to a complete stop. however, with right gear, you could climb that hill at its maximum potential (P=Fv) which would be at max hp of 1.5hp. sure , you could gear it down even further and go slower producing more torque, but all that would do would be to accelerate to a slower speed faster and then top out at max rpm at a slower speed, thus now you back off the throttle and you continue up the hill at a slower speed, and lower HP setting and lower torque requirement.

Answer: The engine lacked sufficient torque to accomplish the climbing of the steep incline. The incline imposed additional loading on the crankshaft, so the engine’s crankshaft steadily lost RPM. As its crankshaft slowed, less A/F mixtures were ignited per second as a result, so less over all energy was provided per second, which further slowed my ascent. It must be realized that “IF” the engine’s torque were sufficient to climb the steep hill in the first place, the engine would have never slowed down. A 2.5 HP combustion engine remedied the issue,
<<<<<<<<<<<<yes, more HP required to accelerate or create the greater rate of work. I get this.
as it produced greater torque and even had enough extra torque to start out from a dead stop up the steep hill. From a dead stop (engine just above idle), it couldn’t be the extra HP yielding the more than adequate acceleration up the hill; it was a direct result of the extra torque provided by the 2.5 HP engine.
>>>>>>> the hp at just above idle would then have more Hp available comparitively than the other 1.5hp engine. yes, I get it, more rear wheel torque was developed then.

Point: A given task requires a given minimal torque to accomplish the task. If less torque is produced, then it makes no difference what the “horsepower rating” is, the task will not be achievable.

>>>>>>>>>I dont quite follow you. Now, If the the hp rating is high enough, the minimal torque can be acheieved. ?????? I dont understand the point.

Regarding your original question, yes it is more clear now. The answers I gave are still the same for everything else. But the fact that its in a higher RPM range and is well geared, both should accelerate the same as long as they ar eboth in their peak power band. The engine with more torque has a wider band, so will pull better from lower rpm. The revver will do all of its useful work at higher rpm.

You are still confused over torque and horsepower. You said I could be way off by looking at only that, this is not the case. As hp is calcualted directly from torque, to it really makes no difference if we use power or torque, the sums will be different but we'd ge tthe same answer.

Power = Torque * angular velocity.

I prefer using torque purely as that is the acutal real world driving mechinsm of the engine. The point you make that torque figures alone are pretty much useless is true, without knowing the engine speed they arent helpful. Using the torque value a trear wheel is a more useful figure from a design perspective, as the size of the wheel has no bearing on the torque, but it will affect RWHP reading and therefore acceleration. So say you calcualte your acceleration times with a 14 inch wheel on a dyno, if you are using RWHP you'd need to re run it as the figure would no longer be valid for the new wheel. By using the engine torque you eleiminate this issue. Although is technically not strictly correct you can say that RWHP determines the torque at the rear wheel, but again this makes it sound like power dictates torque when really its torque dictating power.

You are correct in that to use the engine torque, you'd need the gear ratio and rpm of the engine. This for practical purposes makes the RWHP reading from the dyno more convenient to use, however this also leads people into the false sense that the driving force is power and not torque.

By using Engine T * Gearing * %Losses = Wheel torque
and Using the RWHP

Its making more yes you are correct, its making more power so the forces a the rear wheel will be higher. This is only becuase the torque is being applied more rapidly.

Im going to have to dig out my books on integrating curves. I'll have a read up tonight.

So summerise: both ways will give you the same answer. Using the power and vehichle speed is more convenient to calcualte quickly. but using the engine torque is a more powerful tool for analysing engine performance.

Your responces to Gnosis' questions show that you've not changed your thought process about this one bit.

"I know that the force is makeing the acceleration, but as far as what causes the force, isnt that caused by the energy source and its potential energy? (a 75amp/second 10v battery that produces 1hp-second or 750watt-seconds? Ill tie back into this below.)"

the force at the contact patch is caused by the torque on the wheels NOT THE POWER. Power is a way of visualising what the torque is doing. This is why it is more of an abstract concept albeit one that people are more familiar with. Caused by the torque deliviered by the engine, from burning the fuel. so talking about heat into the engine can also be used, but the trype of work that the engine does is torque. and this is the crucial thing you must understand.

everything else is an application of that torque generated.

If you only had rolling road data without rpm signal, you'd be forced to use the power method. This is perfectly valid.

Now we've got that sorted we can look at the practical side of this. Lets talk about a track with a slow hairpin onto a midlength straight. Both cars have 4 gears, one prodeuces 50 Nm more torque than the other. You could gear the cars differently to give the same low down acceleration from the corner. The car with the lower torque would need to be geared shorter to allow it to be running in the higher rpm range.

As you have a shorter gear ratio in first and the same ratio of ratios (if you know what i mean) so that the acceleration performance ws the same the higher torque car would have less top speed.

However! If you geared the cars for the same speed in each gear, the lower torque car would likely fall off its peak power band, and would accelerate slower away from the corner, but begin to catch the higher torque car down the straight.

This is not a problem if you had a lower torque engine with a 6 speed box, as you could gear for lower accelration and still maintain top speed.

you are making the classic mistake that lots of people have made for race engines. They tune their horsepower and gears so it looks perfectly on paper (fom the dyno), but this wont necessarily transfer well to the tack.

That sounds exactly on the button for the physics terminoligy for what is going on. .

As a performance indicator are you thinking of inegrating for an acceptable power band. So say find the area between 80% max power?

I'm still a little confused as why what you mean when you say the horse power curves are the same. As that to me means the torque must be the same as the graphs overlay perfectly. Do you mean that when the x axis is scaled, the curves have the same trend?


Also you mention getting a broader HP curve, so you sacrifice peak HP for gains along more of the rev range. (in effect maximising the area under the power curve). This is essentially the same as altering the torque curve to sustain at higher rpm (I think). My friends is acutally better than me at this, i'll ask him about it tonight.

I'm really enjoying this discussion, its making me think quite hard and polish up on stuff i've not read in quite a while.

I get it, its getting to be less and less of a chicken and egg thing. force is what causes the acceleration. got it. Hp is a measure of of the rate the force does the work. HP is a vision into what the pontential force is though right?

Mk

Just when I think you're maybe finally getting it, you go right back to that association you have with horsepower and force as per your last sentence above.

Horsepower isn't a "vision" or "insight" into the potential force. All force in the combustion engine is strictly produced by torque and torque alone. Torque IS measureable on a dyno. Horsepower is merely a rating of how much "WORK" (not force) can potentially be accomplished per a given RPM via the available torque of the engine. "Work" and "force" are two entirely different quantities just as horsepower and torque are entirely different.

Horsepower should only be thought of as a rating, as in a "horsepower rating", a work rating.

Try to grasp the concept of “horsepower rating” from this next example:

YOU could potentially twist a ratchet (thereby apply some measure of torque) all day long. This also means we could actually give you a horsepower rating of some kind rated per an entire day’s work. Let’s assume that the absolute torque limit that you can apply per your ratchet is 200 ft pounds and the conveyor belt you’ve been manually cranking with your ratchet has only required 20 ft lbs (piece of cake). This conveyor belt is carrying buckets of material to the top of a silo, then dumping the contents. Suddenly, someone accidentally overloads one of the buckets on the conveyor belt with 400 pounds of material. Doh!

At this point, you HAVEN’T lost your willingness to work neither have you lost your “horsepower rating” (that work which you are capable of accomplishing per day). You simply lack the strength to apply any more than 200 ft pounds of torque to lift the excessive 400 pound load. Since torque is the ONLY factor causing the conveyor belt’s movement and you lack the torque to lift the 400 pound load, all work has ceased. If you stepped over to an identical conveyor belt that didn’t have the excessive 400 pound load, you’d be able to continue at your rated horsepower rating for the rest of the day. Insufficient torque is the issue here, NOT insufficient horsepower.

That's why I stated it WASN’T lack of horsepower that prevented my 1.5 HP mini-bike from climbing that steep incline; it was the engine’s sheer lack of torque, as torque is the ONLY force being produced by a combustion engine and I cannot state this enough. Horsepower is a work rating, torque actually causes the rotation and acceleration of the crankshaft.

I used a 1.5 HP lawnmower engine to power a mini-bike and I geared it to yield a top-speed of 35 MPH while on flat ground with my body weight of 140 pounds. However, while riding at 35 MPH on flat ground, I came to one of the steepest streets in my neighborhood. My mini-bike steadily lost speed and would not make it all the way up to the top of the steep hill.

Your thinking is correct but you have the wording wrong zanick.

Rate of change of kinetic energy wont tell you the force nor acceleration. K.E = .5mv^2

The thing you want is rate of change of momentum. momentum = mv
Dy differentiating this you get the magic formula of F = ma.