Brake Horsepower and Shaft Horsepower

[Terry Newton]

What is the difference between brake and shaft horsepower?

I believe all automotive engines are rated for brake horsepower. However, I think all non-ducted fan aircraft engines are rated for shaft horsepower.

The shaft hp makes sense b/c a shaft sticks out of the motor, and this is where your power is going. I presume you measure the peak HP at your shaft.

However, I don't see why you could not mount a shaft on a 230 inline six and get the same rating.

Anyone with a better understanding of this than me?

Are there other units that I don't know about when speaking of horsepower?

[Cheyenne Dave]

I seem to recall a B.H.P. rating...British Horse Power. Maybe their horses are smaller and need a different rating!

[MoparNorm]

It's early in the morning, so I'm going to pre-qualify my answer as a guess (out of deference to Paul Cook!)
But I would assume that shaft equals HP ratings at the flywheel and brake at the rear wheels? (gross vs. net)
In the early years of the horsepower wars, ferd and gm used flywheel horsepower ratings to get very high (275 to 350 hp) ratings, Dodge has always used rear wheel figures, (200 to 230) which is why a supposedly "underpowered" Mopar motor would always out-perform the other brands. At some point in the late 60's the SAE requested that ALL hp ratings be measured at the rear wheels to provide some uniformity to the wildly varying ratings of the Muscle Cars. ferd and gm ratings then dropped considerably while the Mopar ratings remained the same. A 260 hp Poly 318 will run circles around a gm 350 V-8 for that reason.
Dodge Truck Association (http://www.dodgesweptline.org)

[Gordon Maney]

The terminology Brake Horsepower comes from the use of a friction brake or Prony Brake to measure the horsepower output of an engine. Any engine's output shaft can be used as a point of connection and measurement for horsepower. As far as i can see from having done some research here, I don't see any functional difference between shaft horsepower and brake horse power. Both are defining horsepower as 33,000 ft. lb. of work per minute.

Over the history of automotive marketing, there have been different numbers offered to represent the same set of circumstances. Some figures were derived at the flywheel, others at the rear wheel [which is useable], and others added frictional horsepower to the flywheel rating. Frictional horsepower is that power lost due to internal friction in the engine.

Unless someone comes along with a turbine specific perspective on this shaft horsepower concept, I am going to think that the two terms are effectively the same; both measured at the output shaft, and in the same way.

[Paul Cook]

First, let me defer to Norm. The horsepower race really began when the "low-priced-three" all had V-8's. It was an advertising ploy.

The "flat-head" six cylinder never was a player. And neither were my beloved "flat-fender" POWER WAGONs. Norm's Sweptlines were in the midst of the horsepower thing.

I was still driving a MOPAR flat-head six on into the 1960's and was honestly not interested in horsepower. The following is from this old man's memory.

How many remember the GMC six cylinder? The "Jimmy" was a short lived performance choice following the Ford (Mercury) "flat-head" V-8's after they had been tweaked to the max. The "in-line" picked up some performance image from the Offenhouser engines in the Indy cars. A lot of speed equipment was available for these engines and they were popular with hot rodders.

Then, Oldsmobile's "Rocket" V-8 was the choice. (Pontiac, Buick, and Chevvy still had no V-8's.) The Chrysler "Hemi" was a luxury car option and was not available in numbers that made them affordable for the everyday rodder. Any early "hemi" applications were in the hands of the top levels of the various branches of motor racing. How many remember when the "hemi" finally beat the Hudson in-line six "Hornet" in the Mexican Road Race?

Early in the horsepower race, two terms came into use, "gross" and "net" horsepower.

Gross horsepower was measured with the engine on a test stand with no load producing accessories connected. The coolant was pumped by a separate pump. There was no fan, no generator (alternator), and no fuel pump - nothing for the engine to turn except the crankshaft. Often air intake and exhaust was handled with exotic plumbing to reduce any restrictions to airflow. If memory serves, the flywheel was specially made for the testing and was not suitable for driving a clutch assembly. Test engines were not randomly selected from the assembly line, but were "blueprinted" to production engine specifications.

Then the engine was run at very high revolutions and the maximum, or peak horsepower was measured at the crankshaft. This was the "gross" horsepower and was the basis for the advertised horsepower.

Net horsepower was also measured on a test stand but with all engine accessories in use and, I believe measured at what would have been the equivalent of the transmission input shaft - in other words, with the flywheel/clutch weight also being driven.

http://www.dashlink.com/~txpwrwgn/83900062-DP.jpg

Note that my 1946 WDX has 74 Net Horsepower at 2300 RPM.

We have seen that some POWER WAGON owners have blueprinted their PW engines and added available speed equipment with significant gains in performance.

RPM's equate to horsepower and these blueprinted engines will definitely turn faster than 2300 RPM.

I do believe measuring horsepower at the drive wheels is the best way since it best represents the actual useable horsepower.

Now, here's a philosophical question. Suppose we measured the horsepower available at both the drive axles. Obviously 4WD will reduce the power measured at the rear axle in 2WD, but would it reduce the available horsepower 50%? Would the front axle show only 50% in 4WD or would both axles each measure more than 50%? And, if so, does that mean the same engine/transmission/axle ratio actually develops more real power to the ground in a 4WD than in an identically powered 2WD?

My flat-head, flat-fender brain combination has run out of power on this one. Have fun with it.

[Gordon Maney]

Okay, I will offer my speculation. I think in terms of horsepower at the drive wheels, it would be 50% of the theoretical power available to the rear wheels of the truck in two wheel drive that would be available at either driving axle in four wheel drive, and likely it would be somewhat less than 50%, due to mechanical efficiency issues. There are always losses.

[MoparNorm]

Probably pretty equal, since both axles are post transfer case, ...now how about lockers front and rear ha!ha! The power sent to each axle does change when climbing, as the rear, or loaded axle does most of the work. On trucks with 44 front axles and 60 rear axles, the 60 takes most of the punishment, however climb that same hill in reverse and occasionally you will break a hub or a half shaft in the 44. The only variable is more of the weight was transfered to the 44 as the front, now rear when in reverse, was forced to carry more of the load.
MN
61-71 Dodge Truck Association (http://www.dodgesweptline.org)

[Gordon Maney]

Now you are introducing other more tangible, and less theoretical or abstract dynamics.

Torque is developed under load, as is horsepower. The average, uninformed person thinks that an engine with a rated torque and horsepower is sitting there manifesting all of that specified performance, even at idle.

So, yes, you could and would have one axle under greater load than another.

[S.Swanson]

Several random thoughts:

Torque is a measurement of work/time production.

Horsepower is a function of torque and time (torque x rpm / 5252).

Gear reduction multiplies torque yet preserves horsepower, frictional loses aside.

Power division doesn't reduce torque or horsepower, again frictional losses aside.

It would seem to me that, given any engine rpm/load, the same horsepower measurement will be found at any motive wheel, assuming a direct power transfer to that wheel, even if that power transfer is shared by 3 other wheels.

Standing by for correction.

Steve

[Gordon Maney]

Torque is defined as force applied at a radius. Torque can be applied and no work can be performed, if no rotation occurs. Consider the frozen bolt. We apply torque, it does not turn, so there is no work, even though you might be tired.

Torque is entirely unrelated to time or work or motion. A force of one pound applied at a radius of one foot yields one pound foot of torque. Frequently the incorrect term foot pound is used for torque; it seems as acceptable as the word aint.

Work is defined as a weight moved through a distance. One pound of weight moved through a distance of one foot yields one foot pound of work. There is still no time dynamic in work. You could take all day to move the weight, or move it in one minute; it would be the same amount of work.

Time enters in when you consider power. It takes one horsepower to perform 33,000 foot pounds of work in one minute. I can move as much gravel as a large tractor, but the tractor can do it faster, because it is capable of developing more power.

I agree with your frictional losses aside comments, but.... the realities of drive train and power transmission devices are such that they cannot ignore frictional losses, even if they are minor.

[MoparNorm]

...that a 200 net HP motor would be applying 200 at each wheel, assuming there is no friction loss???? Is that like power in parallel (sp) vs. power in series??? ha!ha!
MN

[Gordon Maney]

No, it would be 100 HP at each of two wheels.

You will notice that colored hi-liters of certain brands, based upon the way the bodies and caps are designed, can have one hi-liter stuck onto the end of another. If you connect several of them together, end to end, it increases the voltage and makes the color brighter.

[S.Swanson]

Oops.

Steve

[S.Swanson]

Originally posted by MoparNorm ...that a 200 net HP motor would be applying 200 at each wheel, assuming there is no friction loss???? Is that like power in parallel (sp) vs. power in series??? ha!ha!
MN

Originally posted by Gordon Maney No, it would be 100 HP at each of two wheels.

I don't see how this could be. I guess I'm going to have to make a call to my mechanical engineer friend and have him 'splain this to me. Until then, bear with me here and help me understand where I'm off.

Let's imagine that we have a truck suspended in the air, with an engine producing 400 lb/ft of torque @ 2626 rpm, turning a transmission at a 1:1 ratio with full lockers in each differential at 4:1 ratio. This gives us 400 lb/ft torque/200 horsepower at the crank, and 1600 lb/ft of torque/200 horsepower at each axle shaft.

Now Gordon - here we're going to have to ignore frictional losses as we have no idea what they are, but I would suggest that they could be consistent, or very near consistent, at both differentials. Additionally, we're going to have to ignore the tire radius and its effect on the torque measurement at the ground, since we also don't know what tire size we're working with.

With our truck in the air, engine happily spinning away, if we measure the torque at any of the 4 axle shafts, we will measure 1600 lb/ft of torque and the rpm at 656.5 giving us 200 horsepower since any of the 4 axle shafts is representing the power output of the engine.

This does not mean that our truck is producing 800 horsepower - it means that it's producing 200 horsepower at 4 different locations.

Yes, some of the torque will be consumed in transferring it 90 degrees from the propeller shaft to the axle shafts. Again, for simplicity's sake, let's assume that we're using exactly the same differentials front and rear and they have the same frictional inefficiency. The losses will be consistent to both axle groups resulting in the same torque applied to all 4 axle shafts.

What am I missing?

Steve

[Gordon Maney]

I think to bring us together on this, if the truck was performing work at a rate made possible by the presence and availability of 200 hp at the drive wheels, that would be the result of the combined efforts of whatever number of driving wheels we had. So, if it was in two wheel drive, the combined effort of the two driving wheels would be working together to deliver this 200 HP of power, rather than each wheel producing 200 HP of power. Does that seem reasonable? Crudely, then, each wheel is applying 100 HP to the road’s surface.

Imagine it as the truck as an assembly applying power to the ground; you can view it as two wheels, or just one wide wheel, it hardly matters, but if in theory we have 200 HP available, that is all we have, we don't have it to that magnitude simultaneously at multiple wheels, it is our total. I suppose if we had a locked differential, and only one drive wheel in contact with the road, then, yes, we would have 200 hp available at that wheel, but not under the typical tractive circumstances where there are two driving wheels on the ground.

As far as your remarks about torque and rpm, I think they are extraneous to the point upon which you are choosing to differ. However, it seems your torque assertions are based at least in part upon [reasonable] assumptions about the torque multiplication possibilities offered by a gear reduction setup. OK, that is fine, but keep in mind when you increase the torque at the output gear, you decrease speed at the same rate.

Let's borrow an electrical analogy for the moment: Power is equal on both sides of a transformer. That is to say that horsepower will be approximately equal on both sides of a gearset. Increased torque that has been gained via gear reduction can move a greater load, but at a slower speed. Recall that power is work per time, not just work.

[MoparNorm]

...you're talking about 2 wheels while Steve and I are talking about 4 wheels, in a vacuum, with lockers,...what color Sharpie is brighter???
Besides isn't there resistance in that electrical equasion, and am I the only one here getting a headache?? ha!ha! You MUST be snowed in again today!
Now if we had the new Jeep, fluid drive quadratrac, every bit of that 200 hp (or torque in the other post) could be applied to any wheel at any time, even if the other three wheels were off of the ground or had lost traction.

Actually I'd be curious to know if there would be 200 lbs ft. at each wheel, or only 50 lbs ft. multiplied by 4???? Otherwise wouldn't you have more torque than the sum of the total? (800)
In water pressure we know that even if the volumn increases, pressure does not. 100 lbs pressure in a 1/2" pipe is still 100 lbs pressure in a 1" pipe, but the volumn is increased. Is that what Steve is saying here?
Isn't this fun!
MN

[Cheyenne Dave]

If you guys keep this up, I may actually learn something!

[Gordon Maney]

Two, four, who cares, the total number is the medium or mechanism through which your power is delivered. It is divided up among them.

Personally, I think Green is, and it is my favorite when I am hi-liting critical information.

Yes, there is resistance, and in both worlds, actually. In the mechanical world it is called friction. I find Advil works pretty good; take four, with food.

Pressure in a closed system is equal throughout and exerted at right angles to the confining surfaces. Having said that, everything changes as soon as there is flow. Your garden hose demonstrates that. Pressure way out at the end of a flowing hose is lower than it is at the tap. Flow is a huge dynamic in fluid property discussions.

Norm, you are a nut. hahahahahahaha

I think pretty soon we are going to have to go back to one question at a time.

[MoparNorm]

I understood everything except the one question at a time part....
ha!ha!
MN
<A href="http://www.dodgesweptline.org"> Click Here </A>

= )

[MoparNorm]

Originally posted by Cheyenne Dave
If you guys keep this up, I may actually learn something!

Oh ye of little faith,...don't count on it!!

<A href="http://photos.imageevent.com/moparnorm/dodgesweptlineorg/icons/Holister%20Line-up.jpg">Click Here</A>

[S.Swanson]

Okay I see what Gordon is saying now, but I'm still stumped. So if I take 4 advils and highlight them green, then multiply by the pressure in in the hose, I can...uh, no, I'm still stumped...

And no, I hope nobody thinks I'm suggesting that we could multiply the horsepower output of the engine by the number of axle shafts. I'm only thinking that there is only going to be 200 horsepower produced, in this example, but that that 200 horsepower will be available at any axle shaft at any time. I guess that's where I'm off. (Thanks for your patience guys)

Oh, and I'm using torque and rpm figures because, for me anyway, it puts a face on the question. Maybe I'm alone in this, but I can see numbers more easily than I can see concepts.

Going back to my truck floating in the air, you're saying that if we measure the torque (or horsepower) at two wheels at the same time each axle shaft will be transferring, roughly, half of the total.

See, this is where I wander off I suppose, because I'm still thinking that if there's 400 lb/ft of torque/2626 rpm (200 horsepower) going into the 4:1 differential, we're going to be able to measure 1600 lb/ft of torque/656.5 rpm (200 horsepower) coming out of the differential.

Wait, wait, light coming on. Since horsepower is a measurement of work produced and the work produced is being divided between numerous axles...

Steve

P.S. - Well I think this thread has been thoroughly hijacked... Started out as shaft -vs- brake horsepower.

[MoparNorm]

Originally posted by S.Swanson

P.S. - Well I think this thread has been thoroughly hijacked... Started out as shaft -vs- brake horsepower.

We couldn't answer THAT question so we made up our own....
ha!ha!

<A href="http://photos.imageevent.com/moparnorm/editingalbum/websize/DCP_0489.JPG">Click here to call a cop</A>
MN

[Terry Newton]

Powered by a screaming Banshee! :)

[Cheyenne Dave]

Now that puts a new twist on things...wish I'da thought of it!