moroso calculator

[Stan Weiss]

(.00426 * 125mph)^3 * 3100lbs = 468.0804796875 flywheel hp

1 / .00426 = 234.7417840375587

(125 / 234)^3 * 3100 = 472.546075425212 flywheel hp

Code: Select all

Equation       (125 / constant)^3 * 3100
        Value of constant     Evaluated Equation / HP
         225.000000000000     531.5500685871058
         225.500000000000     528.0220929567211
         226.000000000000     524.5252693019667
         226.500000000000     521.0592545410847
         227.000000000000     517.6237101164451
         227.500000000000     514.2183019251007
         228.000000000000     510.8427002505496
         228.500000000000     507.4965796956964
         229.000000000000     504.1796191169799
         229.500000000000     500.891501559643
         230.000000000000     497.6319141941316
         230.500000000000     494.4005482535905
         231.000000000000     491.1970989724405
         231.500000000000     488.0212655260155
         232.000000000000     484.8727509712371
         232.500000000000     481.7512621883068
         233.000000000000     478.6565098234001
         233.500000000000     475.5882082323365
         234.000000000000     472.546075425212

[Chargermal]

Stan,

Which constant does the Moroso rule use? I understood it was 234 (and I dont have one handy!)

[Stan Weiss]

Sorry but I do not have a Moroso rule. May be someone with one can let us know.

[ijames]

(The formula I derived from physics says average wheel hp = 197*weight/et^3 and plugging in 11 flat and 3500 lbs that gives 518 average wheel hp so if the Moroso says 520 hp it gives wheel hp, not flywheel/engine hp.)

Sorry to be blunt, but where did you get the "197" constant?

I ask because a "constant" for RWHP CANNOT BE DETERMINED, due to varying factors as Widowmaker has explained...

The accepted formulas for FWHP are:


Hale - HP = (mph/234)cubed x weight (lbs)

Fox - HP = (mph/230)cubed x weight (lbs)

Both the 'constants" (234 and 230) were arrived at by simply taking the results of counltess strip runs using the empirical data. (see link below)

The same CAN'T be done for RWHP, becuase the inertial losses vary to a FAR greater extent for every drivetrain. ITs IMPOSSIBLE to generate an accurate "constant".

Go here for a LOT more reading.

http://www.stealth316.com/2-calc-hp-et-mph.htm

Sorry to be slow answering but work, well, it's been work . The 197 in the formula I posted comes from converting kilograms, meters, and watts into pounds, feet, and hp. It does not represent any kind of empirical fudging or correction, just unit conversions. If you know the weight of an object, the distance, the time to cover that distance, and assume constant acceleration, you can derive the formula that I posted. It gives the net power used to accelerate the object, or in other words, the power applied to the track by the driven wheels. It says nothing about drivetrain losses and nothing about engine flywheel hp, it tells how much power must reach the track to get the object to cover that distance in that time. The assumption of constant acceleration sounds poor given how fast a car will accelerate in first gear compared to high gear, but it greatly simplifies the math and gives the lowest possible hp estimate. Any other acceleration assumption will require a higher peak hp because some of the time the accleration and thus the hp will be less, so it is a conservative lower bound. Now, the formula is in terms of et, and we all know that et really depends critically on traction so it is not a very accurate number to plug into the hp formula; the mph is much more reproducible. Again, using constant acceleration the et and mph are inversely related: et = a constant / mph and mph = the same constant / et. I used a bunch of et-mph pairs from a turbo regal website because that's the kind of car I have, and I found that for cars running 7's up to 13.50 the best constant was 1356, and for cars slower than 13.50 a more accurate constant was 1370. You can also derive this constant from phyics and you get something like 1450, which tells us that constant acceleration isn't really that great an assumption but again, it's simple and it ain't bad. The "best" number will be different for every different car shape because we are totally ignoring wind resistance, too. Now, if you take the hp formula I gave, hp = 197 * weight / et^3, and substitute 1362/mph (using 1362 instead of my 1356, a slight tweak I'm sure due to using different cars to get the number) for the et, hp = 197 * weight / (1362/mph)^3, and do the algebra you get hp = (mph/234)^3 * weight which is your "Hale" equation. Using my 1356 would give 233 instead of 234 in the same formula, again because I wanted the most accurate numbers for turbo regals. That also shows that your Hale formula is giving wheel hp and not flywheel hp, by the way. Using this formula does involve an empirical constant, the 1362, whereas the one using et does not, but I think it is a net win given how much easier it is to reproduce mph compared to et. Oh, I guess I should say that the uncertainty using et=1356/mph for the regals was +/- .5 sec or so, and my guess is that for a manual trans car the drivetrain losses are small enough that they are probably buried in the "noise" of trying to make identical runs anyway, so the difference between rwhp and flywheel hp is going to be small enough that the formula could be said to give both or either. However, with an automatic trans the apparent losses will be much greater so that's where it should be obvious that the formula is giving wheel hp and not flywheel hp.

You state that "a "constant" for RWHP CANNOT BE DETERMINED" but I think that in the most practical sense you have it backwards. It is the actual power applied to the track through the tires that moves the car (and that is what we really care about), and we can directly measure the result (the et and mph), so we can directly measure the "net" or driven wheel hp. What we cannot easily get a handle on is all the drivetrain losses, and so we cannot easily or accurately use the wheel hp to calculate the engine or flywheel hp. The way you phrase it, you are saying that we can accurately measure the flywheel hp (which is true, with an engine dyno, but that's not convenient or easy for most people), but that we cannot accurately turn that into rwhp. That may be true in the strictest sense but we can easily get the rwhp so to me it is much more relevant that we cannot accurately and easily turn that rwhp into flywheel hp. That's why I said nothing about flywheel hp in my first post, and that's why I cranked through the physics in the first place to make sure that the formula I was using gave wheel hp and not flywheel hp based on someone's estimate of drivetrain losses.

Oh, hsutton is right that the mph in the formula is the instanteous mph as the car crosses the finish line, and that isn't quite what the time slip gives, but I chose to ignore that . Also, Barry Burch's formula uses 0.0426 which is 1/234.7, probably a result of some roundoff in starting with 1/234, and the 0.0426 need to be cubed as well: (0.0426*mph)^3.

I haven't gone and read the Am J Phys article but most of the info on the link you posted really involves determining the empirical constant in the equation et=constant/mph for different kinds of cars. The basic derivation of hp =197 * weight / et^3 barely qualifies as a college physics quiz question and contains no empirical constants.

[Chargermal]

"ijames" I appreciate the response.

I dont know enough physics to comment on your calculations..but consider this.......

If you know the weight of an object, the distance, the time to cover that distance, and assume constant acceleration, you can derive the formula that I posted.

My comment re - "impossible to predict RWHP" stands true...when you consider you have used "constant acceleration" to arrive at a result. (No criticism here - just pointing out that this in itself is a "fudge factor" - just in another way - its still inaccurate data.)

Yes, I agree your formula stands as useful as any other "predictor"...Moroso included, but of course in reality the only deadly accurate mechanism to establish HP is an engine Dyno.

The nature of "HP Calculators" is such that most use them as "predictors" , long b4 an "ET" is known..

.......BUT, on the positive side, we both know even RWHP dynos are flawed, and at least your formula provides for an interesting comparision with a RWHP to determine how "happy" a RW Dyno might be.

[Stan Weiss]

Any of these methods is only a guess. If I have two engines which make xxx.xx HP @ yyyy RPM but the shape of the curves is different they will accelerate at a different rate. If I have the same engine in two different vehicles and one has a CD 0.28 and a frontal area of 14.35 and the other has a CD 0.62 and a frontal area of 37.5 what do you think we will see?

[sbchevy1]

One will have more mph.

[Chargermal]

the other has a CD 0.62 and a frontal area of 37.5 what do you think we will see?

A drag racing Bulldozer??????

[JBM770]

I know this is a super old post but bear with me.

I want to run 13s or better in a full trim V6 Holden Statesman. I know it may seem slow but its predominately a street car and I will have a limited budget.

The Moroso FWHP is about 345. The Stan Wiess figure is 315. The car weight plus me is 3500 lbs.

I watched this video: to determine the inlet flow I need to get that FWHP.

For 28" of water I need 235 CFM per cylinder with 94% flow differential between the inlet and exhaust to get the 345 HP.

FWHP / 0.26 / No Cyl / Flow Differential = CFM per Inlet (FWHP = CFM per Inlet * 0.26 * Flow Differential * No Cyl)

I know this formula works because a stock head flows 155 / 111 @ .500 and is rated at 170 HP.

Intense Racing has a Stage 3 head that flows 224 / 212 with .500 valve lift.
http://intense-racing.com/Merchant2/mer ... e=3800_HCV

Is it this simple or am I missing something?

If I use a small turbo instead of the Intense Racing head how does this change the calculations or is that a trial and error exercise?


Thanks