prelude_93vtec

hey everyone. i was wondering if theres any way of converting WHP to just plain HP or crank HP? i know stock is 190 hp and i think around 167-175 whp.

just wanted to see how many hp at the crank there really is

dagle

iTrader: (1)

i dont think theres an actual calculator.. because wheel horsepower is just the actual horsepower to the ground. but i think most people will agree that you lose between 10-20% to the wheels..

TylerC161

just figure it with 15% drivetrain loss. Thats pretty standard. So times your numbers by 1.15 and thats crank. Just use a regular calculator.

117

Just to be clear, if you're assuming a 15% drivetrain loss (i.e. 100 HP at the crank = 85 HP at the wheels), then you need to take the WHP figures and multiply by 1.1765. In this case, 85 WHP * 1.1765 = 100.0025 crank HP.

dagle

iTrader: (1)

yeah i was gonna say you dont just multiply by the 15% u took.. u have to do


(100/85) = 1.1765..

or for real life, the simple agebraic equation

(CRANK HORSEPOWER)
__________________________
(% HORSEPOWER REMAINING)

^-- and that figure multiplied by WHP to get BHP.


or to figure WHP from BHP-

BHP * (100% MINUS the power lost) = WHP

prelude_93vtec

still confused?

wait so you calculate whp by 1.17 to get crank?

117

It's basic algegra.

WHP = Crank HP * (1 - driveline loss)

Say you suppose that driveline loss is 15%.

WHP = Crank HP * 0.85

Now solve for Crank HP:

Crank HP = WHP/0.85

Or in other words:

Crank HP = WHP * 1.1765...

prelude_93vtec

ok gotcha

thanks

mgags7

Firstly, I wouldn't take 15% for a standard drivetrain loss #.

Secondly, horsepower is not a physical quantity, it is a function of torque and rpm, so you can't talk about physical drivetrain losses directly to horsepower. It can be argued that losses increase with rpm, either way, I wouldn't try to make any direct connection with hp numbers.

I'd suggest looking at torque losses, and if you notice, stock torque numbers for the h22 (158ft lb torque) versus what most of us put down at peak on stock h22s, and you get something more like a 7% loss.

Just throwing that out there, but my take on this entire topic is that you can't really quantify that loss to a solid # that you can depend on.

prelude_93vtec

ehh i just needed an estimate anyways lol

but ya that is true i think torque would be better to look at but kids in my school probably have no idea wth torque is ll

117

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by mgags7 &raquo;</TD></TR><TR><TD CLASS="quote">Firstly, I wouldn't take 15% for a standard drivetrain loss #.</TD></TR></TABLE>

I wasn't advocating 15% as the magical number. I was using it as an example.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by mgags7 &raquo;</TD></TR><TR><TD CLASS="quote">Secondly, horsepower is not a physical quantity, it is a function of torque and rpm, so you can't talk about physical drivetrain losses directly to horsepower. It can be argued that losses increase with rpm, either way, I wouldn't try to make any direct connection with hp numbers.</TD></TR></TABLE>

HP is very much a physical quantity. HP is a measure of power, i.e. work done per unit time. With a little mathematical manipulation and some very well known equations, you can arrive at the equation HP = Tq * RPM / 5252, as I'm sure you know. Saying HP is not a physical quantity is like saying area isn't a physical quantity because it's L x W for a rectangle.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by mgags7 &raquo;</TD></TR><TR><TD CLASS="quote">I'd suggest looking at torque losses, and if you notice, stock torque numbers for the h22 (158ft lb torque) versus what most of us put down at peak on stock h22s, and you get something more like a 7% loss.</TD></TR></TABLE>

Loss in torque = loss in HP, but how much loss in HP depends on what RPM the torque loss is occuring. Just looking at the formula should tell you that.


Modified by 117 at 9:36 PM 11/18/2007

mgags7

Physically measureable is what I meant to type, though I suppose you could rig something up to take care of that. You have me on that one.

Of course I know that less torque means less hp, the point of that statement was that drivetrain loss isn't going to be as linear as compared to hp as it is to torque, unless I'm seriously mistaken about the magnitude of losses due to the added friction at high rpm.

117

I would think that drivetrain loss would be more linear with respect to torque, as you're taking RPM out of the equation, which eliminates one variable. After thinking about it a bit, I think you're 100% correct on that one. EDIT - see my reply below, I don't believe the preceeding argument is valid.

I don't believe there ever has been a clear cut answer, but most of the things I've read state that it's a combination of a fixed amount of driveline loss combined with a percentage of loss since the value of power (torque) lost will increase with RPM since there is more power (torque) being transferred through the gears (i.e. higher frictional losses).

I'm by far an expert with regard to this subject, but it would be interesting to take some WTQ measurements and WHP measurements and assume some fixed percentage of loss, then compare the two and see what the resulting Crank HP curves look like. It would be very simple to do in Excel with data output from WinPEP.

Oh, not to nitpick, but for the physically measureable part, if you can measure torque and RPM, you can directly measure HP. In a HS physics class many, MANY years ago we figured out HP by timing a person climbing the stairs.

Damn, this is probably the most technical thing I've seen in the Prelude forum in a long time. mgags7!


Modified by 117 at 9:45 PM 11/18/2007

117

Check that. I just did some analysis with some dyno results in Excel. One thing I didn't think of that threw me off is that RPM isn't really variable, as you're measuring TQ and HP at some given RPM, so really RPM is fixed.

If RPM is held fixed, then HP and TQ are then linearly related, and an xx% loss in one equals the exact same percentage loss in the other.

Now the question is, is the percentage fixed at all RPMs. I would say probably not, so that compounds the problem. The only way to know for sure is to engine dyno an H22, then install it and get it on a chassis dyno to get WHP numbers and see how the loss changes with RPM.

gogunkergorilla

You guys are trying to figure this out and there are too many variables that you don't know. Your best bet is to get a dyno of a healthy bone stock prelude and compare to the sae numbers. That should get you a ball park figure. Then there are just things like frictional differences which should be negligible, and wheel weights and tires.

mgags7

It is really tough to model all the factors involved, because we know that different RPM levels should definitely have an effect on losses, all the way from bearing and gear friction in there to the MTF's perfromance at high speeds, whether it is shearing off the gears at high speed from 'wind', etc.

That of course begs the question of, not only is the loss going to change with rpm, but on the countershaft and differential size, it will change with mph as well.

Technical discussion

prelude_93vtec

i just realized.

i did that calculation HP = Tq * RPM / 5252

Horsepower: 190 hp Max Horsepower: 6800 rpm
Torque: 158 ft-lbs Max Torque: 5300 rpm

158*5300 = 837400

829816/5252 = 159.44 = hp

i think vtec messes up that whole calculation since the answer is suppose to be 190

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by gogunkergorilla &raquo;</TD></TR><TR><TD CLASS="quote">You guys are trying to figure this out and there are too many variables that you don't know. Your best bet is to get a dyno of a healthy bone stock prelude and compare to the sae numbers. That should get you a ball park figure. Then there are just things like frictional differences which should be negligible, and wheel weights and tires.</TD></TR></TABLE>

ya that would be cool if someone had a stock lude and dyno'd it.

Rosko

iTrader: (1)

^^ I think you just proved that max hp and max torque don't occur at the same rpm. And the fact that horsepower and torque always cross on a graph at 5252 is gonna give you about the same # anyways when you are doing the calculation at 5300rpm

Rosko

iTrader: (1)

in other words the equation is only going to help you find max hp at a given torque and rpm and vise versa. it can't tell you max horsepower from a given torque because they usually don't occur in the same spot (5300tq and 6800rpm for hp)

prelude_93vtec

gotcha

mgags7

93, maybe you could use some clarification on what HP actually means...just to put some info out there.

Basically, HP is just a measure of the engine's ability to carry torque to high rpm. It is usually the best measure of an engine's performance because it incorporates both of the most important of engine characteristics: Torque output and Powerband.

You can make a ton of torque, but have bad performance because you have a low or tiny powerband. An 18wheeler is a good example of this, as are 2stroke engines, they typically have a very short powerband.

You can also have a very wide powerband, and spin a ton of RPM, and not perform well. cough b16 cough Just because you don't make a lot of torque.

The torque measurement you see on a dyno sheet is the "Average torque" that the engine is outputting at that given time. What it is actually comprised of is the 4 spikes in torque caused by each of the cylinders firing, and in between those pulses there is the line showing the torque used in pushing the next cylinder, on its compression stroke, to TDC, and torque is used in order to do so. Anyways, point of the matter is, you have an average torque number comprised of 4 peaks and 4 valleys on the instantaneous torque graph. Now, think of 200ft-lbs of average torque at 2000rpm, then think of 200ft-lbs of avg torque at 4000rpm. The average is still going to be the same, 200ft-lbs, but at 4000rpm you're applying that torque twice as fast, so double the torque is really being applied to the wheels.

So, more RPM is always better in that scope. The problem is, getting engines to make torque at high rpm is very difficult. It increases the demands on the head and camshafts to efficiently get more and more air in and out, while at low rpm this can be done more easily because the overall volume of air is much less.

That, is why horsepower is such an effective analysis of an engine's performance.

/rant

Rosko

iTrader: (1)

good explanation Matt!

NirVTEC

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by gogunkergorilla &raquo;</TD></TR><TR><TD CLASS="quote">You guys are trying to figure this out and there are too many variables that you don't know. Your best bet is to get a dyno of a healthy bone stock prelude and compare to the sae numbers.</TD></TR></TABLE>

1993 VTEC
Stock H22A1-160k miles burning little oil
155whp/135wtq on my local dyno
I paid for a baseline for this kid just to have the stock number.
Average i/h/e/vafc=175whp/143wtq(173-178whp & 141-145wtq range)

prelude_93vtec

thanks clint