Horsepower, what is it?

95 shaw,

Exactly!

A very simple scenario based on some rough 2-cycle numbers: Let's say a Yamaha KT100 produces peak torque at 10,000rpm. It produces peak horsepower at 11,500rpm. RPM goes UP more than torque goes DOWN, or peak horsepower would not be happening at a higher rpm than peak torque! It really is that simple.

As you say, if we gear both scenarios to the same axle rpm, more horsepower at the crank produces more horsepower AND torque at the axle (since axle rpm is the same).

PM
 
Let me try this. Let's say my instrumentation only puts out a torque number. I have no idea what the RPM is so I can't calculate an HP number. I have another gauge hooked up to the axle, and it can read both torque and RPM. Now I can calculate the HP, "at the axle". There's a clutch on the engine, and the clutch is holding the engine at some RPM, I don't know what RPM, the gauge will not tell me, as I said before. I'm hold the brake so the axle won't turn, I can't calculate the HP at the rear axle because there's no RPM. I let off the brake, the engine starts gaining RPM, but in this hypothetical engine, the torque remains the same irrespective of the RPM. I still can't calculate HP at the engine because there's still no RPM reading. I have no idea what the horsepower is at the engine. Now the axle, that's a different story. Because I have a reading of both torque and RPM, at the axle, I can calculate the HP. I don't care what the HP is at the engine, all I know is it's putting out the same amount of torque as the axle gains in RPM. That means I can calculate the HP at the axle, and the faster it turns, because the torque stays the same, the higher the calculated HP number
goes.

The point being, the axle has no idea what the calculated HP number is at the engine, and doesn't care, of course the HP is going up at the engine, I know that, but what difference does it make when you get back to the axle, all the axle sees is a fixed amount of torque, and, as the axle continues gaining RPM, I would calculate an increase in HP.

To oversimplify it; HP is the calculation of how much work is getting done in a given amount of time. More work, less time, more HP. The inverse, of all 3, same thing. (Less work, more time, less HP.)
 
Now your reversing the scenario . Same applies if you know the gear ratio and axle rpm ; you known the engine rpm thus you can calculate engine horsepower .
Equally the inverse is true . Knowing engine rpm and torque along with gear ratio .
Axle rpm can be calculated .
 
Now your reversing the scenario . Same applies if you know the gear ratio and axle rpm ; you known the engine rpm thus you can calculate engine horsepower .
Equally the inverse is true . Knowing engine rpm and torque along with gear ratio .
Axle rpm can be calculated .
I don't know this for a fact, and I apologize in advance if I'm wrong, but I think you totally missed the point of my post.
 
Al,

In your original post I/we replied to, you said:
The ideal situation would be to have an infinitely variable ratio transmission so you could keep the engine at peak torque as you accelerated.

My reply simply (and I thought: clearly) pointed out that with an "infinitely variable transmission", running the engine at peak horsepower (not peak torque) would generate more torque AND horsepower at the axle.

Now you have introduced clutch slip to this conversation, which adds many variables.

From everything I've read in this forum (and we ARE in a LTO/oval racing forum, for the most part), the only time a clutch comes into play seems to be on the jump at the start... and only if it's a slow start. I believe the rest of the time clutches are locked and maximum momentum and least possible scrubbing of speed in the corners is the focus. Plenty of guys here will hopefully correct me if I'm reading this wrong.

PM
 
I wonder, if, while watching an engine being tested on a dyno, if people understand that that HP number they see on the screen is the "sum" of a calculation, not a measurement? The dyno is "measuring" 2 things, torque and RPM. The dyno computer computes a number using those 2 components, it divides the sum of that calculation by 5252.1 and displays, on the screen, the results of that calculation under the heading "Horsepower". Very "Simply?" put, a measured force (torque) during a measured length of time (RPM).
"Civil responses" are welcome if you disagree.
 
In your estimation , what is the significance of this number?

It must have some value, or it would not be included. We would just look at torque @RPM otherwise.

I'm not sure I've seen an uncivil post in this thread.
 
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Doesn't an inertia dyno measure the amount of time a motor takes to accelerate a known weight from one rpm to a higher rpm?
If so, what would that be?
 
In your estimation , what is the significance of this number?

It must have some value, or it would not be included. We would just look at torque @RPM otherwise.

I'm not sure I've seen an uncivil post in this thread.
I think it allows us to see if, using a smaller shovel (less torque), shovelling faster (more revolutions), will yield a larger pile of sand (more work) in a given amount of time.

The point where this begins to be not possible is at peak horsepower.
 
Is it? Once the hp line crosses over the torque line to get more work out of an engine hp must increase exponentially over the decrease in torque.
Does It mean gear bound could happen either because torque decreased too rapidly or hp did not increase fast enough?

I'm thinking since hp is a calculation you could graph a line either per torque or hp for the rpm at which gear bound can occur per any engine.
And I think the graphed gear bound line can be either above, below or at maximum rpm generally obtained.

yep, proly wrong dumb thunkin butt if you knew the line for your particular engine you could better guess on getting more usable work out of the engine by either addressing a torque increase or actually a hp decrease.

Might it give a reason to build towards either a hp increase or decrease or torque increase.

I'm thinking it would always be good if you output was an increase in torque.
Needed calculated hp would then depend on rpm and how it relates to torque.

I'm thinking for top end performance you either have to delay the decrease of torque or extend the time it takes for torque to drop or exponentially increase hp . and lots of ????????????

thanks Shaw this is and an education to think about no matter if your thinking is right or wrong. ... :)
 
In your estimation , what is the significance of this number?
It's actually fairly easy, I've done it (everybody knows my love of spreadsheets) with a list of HP at a given RPM, you can calculate the torque required to produce those HP numbers.
 
Without going into a lot of detail, horsepower is a simple way of telling people how much work a machine can do. The original concept was first formulated by James Watt (see Wikipedia) in the late 1800s, in order to describe to farmers, who knew about horses, how much work, compared to horses, his new steam engine powered harvester was capable of. This story is kind of interesting, you should read it.
 
Doesn't an inertia dyno measure the amount of time a motor takes to accelerate a known weight from one rpm to a higher rpm?
If so, what would that be?

Absolutely Bob -- once the moment of inertia is known (of all the stuff being accelerated), getting an "average horsepower" over a measured period of time is totally doable.

PM
 
Is it? Once the hp line crosses over the torque line to get more work out of an engine hp must increase exponentially over the decrease in torque.
Forgive me for sticking my nose into this discussion, but! Torque x RPM / 5252.1 = HP.

10 X 5252.1 / 5252.1 = 10
the numbers 5252.1 cancel each other, so torque equals horsepower. That's why the horsepower and torque curve always cross at 5252.1.
33,000 Lbs. Ft. / 2Pi() radians = 5252.1
Horsepower = a rate of work.
lifting 550 lbs. 1 ft in 1 sec. is defined as working at the rate of one horsepower.
550 lbs. ft sec. x 60 seconds = 33,000 lbs. ft. The work performed in one minute working at the rate of one horsepower.

The fact that these 2 numbers cross, at that point on the graph, has nothing to do with anything else
 
Thinking about gearbound and how it could be simulated in a controlled environment.
Thinking it would be very difficult on a steady state dyno.

Could it possibly be done on an inertia dyno?
 
Time for math class .
18,000 lb load lifting 100 feet in 25 seconds f×d÷t = power
I get 72,000ft lbs with a 6/1 ratio 12000 then divide by 550 to reach required hp as 21.81 . Which seems incorrect as we used a 6 cylinder duetz diesel hydraulic winch to do said job . Our concern was always line pull , don't want to seperate the cable .
 
Thinking about gearbound and how it could be simulated in a controlled environment.
Thinking it would be very difficult on a steady state dyno.

Could it possibly be done on an inertia dyno?
You could definatly map the acceleration rate .
If you used a defind , test time factor that might work .
 
Time for math class .
18,000 lb load lifting 100 feet in 25 seconds f×d÷t = power
I get 72,000ft lbs with a 6/1 ratio 12000 then divide by 550 to reach required hp as 21.81 . Which seems incorrect as we used a 6 cylinder duetz diesel hydraulic winch to do said job . Our concern was always line pull , don't want to seperate the cable .
Line pull.
so, using 6 part line block puts line pull at 3000 lb
3000 lb moving 600 feet in 25 sec
720000 ft/lbs/sec
equals 130.91 hp

This is the hp requirement at the winch cable interface.

Same as single part line load. Better be some healthy cable. 7/8 if I recall correctly for overhead lifting.

6 to one drive ratio makes it 21.8 hp at drive motor. For the bottom layer of cable on the drum. Takes more as the cable layers up.
Safety factor, hyd losses, etc add 100% minimum.

Probably did not need peak hp to move load, just needed the speed afforded by revving the engine.

Yes, I have seen a crane or two.
 
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Yep the throttle was a speed variable once you got started .
Well i had some math parts correct . Not much .
Thanks
And too add : yes you had to input either more throttle or more hydraulic pressure at the top of the drum depending on which you were using to control the lift . Usually throttle .
 
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