So what is HP and what is torque and what is their relation to each other and why do we care? Even if you don’t like technical matters, this one is very interesting because it answers non-technical questions. Instead of me using mathematical equations and talk 100% in technical ways, I will write about reality. Reality everyone understands.

What people say:

I am sure you have heard of people saying things like: "there is nothing like torque" or "this engine pulls hard even in 5th from 25mph", or "when I get to 6000 rpm, that is when I feel the power". Why do people say all that?

In reality:

Anyone who has driven a WRX admits that, if we exclude the turbo lag, the car feels really fast. And that is fine. 0-60mph in 5.8 is pretty fast in anyone’s book.

On the other hand, anyone who has test driven an S2000 usually comes out of the car and says "where is the torque?…this car feels like nothing special under 6000 rpm, why don’t I feel the power?"

Hmmm…something is going on here. And does it have to do with HP, torque, or both?

For simplicity’s sake let us assume that both cars are of the same weight and also use the same exact gear ratios in the tranny as well as the differential/s. That way, we are concentrating on the engine only.

I am going to use a simple equation and 2 dyno graphs.

This is the graph for the S2000:



And this is the graph for the WRX:



Both graphs show HP on the wheels of course. The dotted line in the WRX graph show the curves on a stock WRX and these are the curves that we care about.

Torque is basically TWISTING FORCE. You have to have a rod, or something, turning to produce or measure torque. Where power is "how often" that torque appears which is represented by how often does that rod turn. Multiplying the 2, you get power. Multiplying the 2 and dividing by a constant, you get horsepower.

In an equation it is expressed as follows:

HP= (Torque * r.p.m.) / 5252

That equation says that Horsepower equals with: the amount of torque, times the RPM that torque appears at, divided by this constant number, 5252.

What the hell does that mean and why do I care, I can hear your thoughts of some of you.

Well, let us see how is that represented in those 2 graphs above:

Let us look at the S2000 first. The green curve is that of torque and the red is that of HP. At the bottom/horizontal axis we have RPM and on the vertical axis we have torque, as well as HP. Torque is measured in lb-ft here in the USA, in Europe they like Nm (Newton-meters) or Kgfm. It does not matter how you measure this. The bottom line is the same here. Watch now:

Looking at the green/torque curve and looking at where this curve is at, at 2000 rpm, we are trying to figure out how much torque we got on the vertical axis. In this case, we have about 85 lb-ft of torque.

Hmmm…what does that mean?

Nothing yet. Everything in mechanical engineering and even math is comparable/subjective. A number means nothing in itself. You have to compare it to another number to see what that numbers really means. Is it a large number, a small number? Well, let’s see as we go further.

OK, so now that we know the torque AND the RPM that torque appears at, guess what! We can figure out the HP!

So, (85*2000)/5252 = 32.4 HP

Thus, the S2000’s engine makes 32.4 HP when it turns at 2000 RPM. Excellent. Now we are getting somewhere.

And if you look at the other curve, the red one, you will see that, that is what it is showing us. About 32 HP. Well, that is because HP and Torque are totally related when you express them in such a way. In a graph with 2 axis showing RPM in one and HP/Torque on the other. Specifically, dynamometers do not measure HP. They measure torque and derive the HP curve from the above equation. Guess why: Because it measures, whatever it measures, on the wheels. And the wheels are turning. And as I wrote, torque appears only when something is turning. It is twisting force.

Fine, fine, fine. Let’s zoom out and go to the power curve of the WRX. Doing exactly the same thing we did before for the S2000, we find that the WRX’s engine makes about 115 lb-ft at 2000 rpm. Or 44 HP.

So, 32 HP for the Honda, 44 HP for the Subaru. The Subaru makes 35% more power at 2000 rpm!

Let us look at other points on the curves:

WRX

2000 RPM= 115 lb-ft = 44 HP
2500 RPM = 120 lb-ft = 57 HP
3000 RPM = 150 lb-ft = 86 HP
3500 RPM = 170 lb-ft = 113 HP
4000 RPM = 185 lb-ft = 141 HP
5000 RPM = 200 lb-ft = 190 HP
6000 RPM = 190 lb-ft = 217 HP
6500 RPM = 170 lb-ft = 210 HP

S2000

2000 RPM = 85 lb-ft = 32 HP
2500 RPM = 100 lb-ft = 48 HP
3000 RPM = 117 lb-ft = 67 HP
3500 RPM = 117 lb-ft = 78 HP
4000 RPM = 115 lb-ft = 88 HP
5000 RPM = 217 lb-ft = 111 HP
6000 RPM = 123 lb-ft = 141 HP
6500 RPM = 130 lb-ft = 161 HP
7000 RPM = 129 lb-ft = 172 HP
8000 RPM = 125 lb-ft = 190 HP
8250 RPM = 124 lb-ft = 195 HP

WOW!

Look at the difference.

Here is why the car with the highest output of torque at low and mid RPM is easier to drive and why you don’t need to be shifting all the time! You simply, have enough power! Horsepower in this case. So when you have large amounts of torque at low rpm, you have large amounts of HP at that rpm range. Simple enough.

The truth is that most of us, most of the time, find ourselves between 2500 and 4000 rpm. In that range, the WRX makes from 57 HP to 141 HP. The poor S2000 starts at 48 HP and stops at 88 HP at that same range. Their difference at 4000 rpm is 61% more for the WRX!!! And at 3500 RPM the difference is even larger. 69% !!!

In a very simplistic analogy, if the 2 liter engine of the WRX were 69% larger, it would have been measuring of 3380 cc. Almost 3.4 liters. Just referencing.

The lesson of all that is that:

IN ORDER TO UNDRESTAND THE REAL OUTPUT DIFFERENCE OF AN ENGINE (at every-day-life RPM ranges), WE MUST CONVERT THE TORQUE IN HP (at those every-day-life RPM ranges).

I will stop here for now.

 

The graph of the WRX is for the BHP and the crank and it's not a Dyno graph ( Dynot Jet).

WRX makes a 225 BHP at the crank not at the wheel.

Your analysis is correct, your data set is flaud.

 

Flaud indeed!

Good write-up though, very original.

 

Let us pretend that the data is at the wheels for the WRX as well. The point still stands. Plus I always suspected that the WRC makes a lot more than 227 HP which it is rated at. These acceleration times suggest that. In any case, as I said, it does not matter. Let us assume both are at the wheels.

 

Thanks.

For the record though and since the word has been used twice by 2 different people, I am assuming that you guys are trying to write "flawed". Right?

 

Gravil excellent analysis!

 

Thanks Zapata.

I'd like to expand it though. After the initial posts from members, I have a few thoughts of mine that I wonna pass by you all to see if I am thinking of them right.

 

I am somewhat new to the topic, but tell me what you think of my thoughts on the matter. So, at a given RPM, there is a 1 to 1 mapping of Torque to HP. IE, given a torque figure, there is one HP figure that corresponds to it (by calculation) at a given RPM. So talking about torque vs. HP at a particular RPM doesn't mean much. It seems to me that the topic in which HP begins to shine is when you are discussing an engine's capabilities in entirety. For example, saying the engine can make 200 ft/lbs doesn't mean much, since it could make that at 2000 rpm and die off fast. BUT saying an engine makes 260 HP implies that the torque curve holds on long enough through out the RPM range to get to a point where torque *(rpm/5252) = 260. Thus saying the engine makes 260 hp tells you a little bit more (though by no means all) about the torque curve along with the maximum torque, all in one number as opposed to a graph. Sound kinda right?

 

not to go too geeky here, but it would seem the absolute best indicator of an engine's capabilities would be the integral of the torque curve. That would help promote engine's who's curves are flatter as opposed to those that peak fast and die off.

 

Yep, Flawed not flaud...sorry for that.

 

i think Mr. PULL T'S or as i like to call him Mr.T or sometimes B.A. comment "flaud indeed..." was a directed at the flaw in flaud....now B.A. be nice and drink your milk...
"I dont want any Milk ...and i aint goona fly..."

 

I think, yes, it does sound about right. To add to your initial comment, yes, there is always ONE point where Torque and HP meet. And that is always at 5252 RPM. In addition to that, all engines make more torque than HP below 5252 rpm and more HP than torque above that rpm point.

 

Exactly correct. So who knows how we can figure this out?

In other words, we are looking of an equation or a function that will help us measure the aggregate of all points under those 2 curves. Where is that math teacher of mine in high school who when he was explaining all that, I was fooling around with who knows what...

 

To make it simple, you could just sum the values at each 500 RPM increment, or every hundred to be more precise, to get an idea of percent differences.

That is a reasonable method to get a general idea.

Here is another research project, why is it 5252 RPM. There is a specific answer to this...

 

According to some quick number I ran based on the numbers shown above, the WRX engine only make about 82% of the power of the S2000 engine.

 

1 ft lb of torque is required to move a 1 lb weight 1 full revolution around a 1 ft radius circle, the circumference of which works out to ~6.28 ft (approximately - it's 2 x pi x radius). so that would be 6.28 ft lbs of work performed for each revolution.

1 horsepower is 33,000 ft pounds of work performed in 1 minute.

33,000 ft lbs/min divided by 6.28 ft lbs = ~5252 rpm

 

I was actually going for the concept of there being 2-pi radians in a revolution. Convert revolutions per minute to radians per second:

Multiply RPM by (2-pi/60) = 0.10472 radians per second.
HP = 550 foot-pounds per second
550 / 0.10472 = 5252.1

But your are just taking it from the other approach in relating to Minutes versus Seconds. Both are correct...

 

1. When comparing those graphs, it would be important to know the gear ratio used for the dyno. The type of dyno is also important...

2. Someone also mentioned that it was important to know the area under the curve -- just for "info" purposes, it's rather important and you can just use simple approximation method.

Want a poor mans integrator -- get some engineering or other graph paper (chose the grid of your choice). Stick the graph paper into your printer (in lieu of "plain paper"), and count the squares under the curve. If you want to compare areas, subsect the regions (per Scalbert’s suggestion) by drawing vertical lines at evenly spaced intervals – like every few grid squares..

3. When you start talking CVT, a high-revving peaky engine can be a brute and kick the butt of a torque monster. Gear ratios are why torque AND hp are important...

 

Thank you Capt. Obtuse :P

 

5252 is the number that converts power to SAE NET Horsepower. It is the translator.

 

And I am guessing that, that is the aggregate power under the HP curve. And I am again guessing that, that is so because the S2000 engine goes all the way to 9000 rpm. Am I right?

 

Can you elaborate a little more on this?

 

Start with a 4-speed gearbox (a bit outdated for sure).

The acceleration in each gear is a function of the following:

The torque curve is equivalent to the acceleration in a given gear. (low gear/first gear == high acceleration high gear/forth gear == lower acceleration)

So, if I draw a horizontal line for a torque curve from 2K to 4K, the engine will have acceleration that is also flat from 2K to 4K.

As, the gear changes, the "effective torque" that is seen at the wheels is a function of gear ratio.

AND

If the max RPM of the “mythical toque monster is only 4K”, I can’t increase effective torque. For example, if two engines put out 100 lb-ft at 10K and the other at 5K, the one with the 100 lb-ft at 10K is making more HP and can be geared down to make MORE torque and more acceleration at the wheels..

AND

The acceleration of the car will decrease as I move from 1st to 2nd, 3rd, and 4th. The acceleration force would be like a down-going staircase.

So, if I have a 2:1 ratio (2-turns on the input shaft, for 1-turn on the output shaft, I have cut the revs in 1/2, but increased the torque x 2). It is just the same as an electrical transformer, lever, or pulleys (and the list goes on).

So, the higher the engine speeds, the bigger the gear down, and the more effective torque at the wheels. (Hey, if I had a 1000:1 gearbox it would be able to move buildings, but it sure wouldn't turn a wheel very fast!)

So, now assume that I have this "nasty" little high revving engine that runs at 10K rpm (the torque peak is there and the hp peak is there). Below 8K the engine puts out average torque, but its nothing to brag about. (This compared to the brute engine with flat-line torque from 2k to 4k). (For the peaky engine, assume the torque is flat from 8k to 10k -- that's peaky in my book...)

Now, assume you install a "wide ratio" CVT that keeps the engine on the torque peak at 10K rpm:

Since I now am revving at 10K rpm, I can gear down and the torque that is being put out by the engine is going to be multiplied by something like 10k/4k = 2.5x more than the torque monster. And, in lieu of having "fixed" ratios that produce the acceleration curve that looks like a staircase, the curve looks like a downward sloping diagonal line.


I'm going to include a graph of G-force vs. MPH, and the diagonal line is the CVT and the stair case is the 4-speed box (Sorry, no aero resistance or drag forces are shown -- and I'm running out of time and this is just a graphic demonstration of how the area of acceleration is greater for the CVT curve (diagonal line) vs, that the staircase. So, now assume you have a peaky engine that only makes good power from 8 to 10K, and you can move that diagonal line up twice as high.)

Caveat: there is one area where there is a "problem" and that is the initial "get it off the line" -- I didn't show the advantage that the "torque monster" would have right off the line. (The model for the CVT isn’t easy…)

The basic point is 1) continuously variable gearing negating the need for "flat" torque (when not right off the line) 2) high hp at high speed means high effective torque from "gearing"...

When I get more time, I can compare actual #s (one mythical engine against another...)

Picture of "SAME engine" red line is CVT and staircase is 4-speed.
Vert axis Gs, Horz axis mph

 

Eric, I think I understand what you are saying here but how realistic is such a thing?

I mean what about issues like:

1. Noise. We all know that engines tend to by noisier at high rpm
2. Gas Consumption. Running at such high rpm is this not going to be an issue?
3. Durability. Running at such high RPM for most of the time, wont that hurt durability of that engine and other components? I mean the faster parts run, the less they "live".

 

1. Noise -- yes...

2. Gas consumption would be related to a multitude of factors, but the increasing piston speed is not generally good for gas mileage. (Hey, in "eco'cars" they generally use lean burn and they aren't built for high-revving...)

3. Sure. Outside of some "technical" innovation, it could be messy. (In general, high revs more piston travel per miles traveled)

(About #3 -- I remember when 7K was considered high for a street car and 8K was quite novel -- now look... so, it depends...)

 

 

I only took it to 8500 in the example (there is another 400 revs to go before redline), but yes that is the point. It is the useable engine power.

 

BEFORE:


AFTER:
Williams-Renault CVT Test Car banned from F1

 

Now here is a personal thought. The gurus of mechanical engineering please let me know how wrong or right I am about this.

I am thinking that the other reason why the WRX feels so much faster is the following:

Let us assume that both engines increase their speed at the same time. So acceleration of RPM is the same throughout the RPM band.

We all know about the basics of inertia. So when we are at 3000 rpm with the WRX and the engine makes 86 HP (or, is "pulling us" by a factor of 86HP for that matter) AND we put the pedal to the metal until 5000 RPM, the engine now is "pulling us" with power equal to 190 HP. The difference is 121% and it took 2000 RPM to happen (whatever that means in time/seconds does not matter because we have agreed that both engines increase RPM at the same rate).

Now in the case of the S2000, the numbers accordingly are 67 and 111. The difference is now 65%.

Thus (I am thinking):

THE ACCELERATION WE WILL FEEL, DUE TO INERTIA, AS A DRIVER IN THE WRX IS ALMOST DOUBLE THAT OF THE HONDA.

About the thought that, "but the Honda will give you better numbers if you measure the same thing from 6000 to 8000", I have this to say:

When an engine increases its RPM from 2000 to 4000, that equals to 100% "benefit". When it does from 4000 to 6000 rpm, that is only 50% benefit. And from 6000 to 8000 rpm, that is 33.3% benefit.

So as the RPM range increase get higher, the benefit we get decreases.

WHere am I wrong?

 

Great explanation EricL. That clears up a lot. Just to make sure i understand, you're basically saying that a CVT could manage to keep the engine right on its peak torque, and thus achieve optimal acceleration at all times other than the start? On another note, if you're in the mood to explain something else, how do they actually implement a CVT? I feel like i've read it somewhere but for the life of me i can't remember - and what cares currently have one? thanks.

 

MyBigBlack S:


I simplified a bit and just mentioned that the hp and torque peaks are at the same HIGH RPM.

A CTV can control the engine speed (drive by wire) and gear ratio, so whatever produced the maximum torque and the road wheels will determine where the “controller” will try to maximize the torque to the driven wheels (acceleration). (Keep the engine on its power peak)






http://www.cke-tech.com/cvpst.htm
http://www.insightcentral.net/encvt.html

 

Do a search on caranddriver.com for a review of the Audi A4 3.0 with a CVT. It's an enlightening article.