Just got back from Dallas Mustang Dyno pull

HP= Tq(RPM)/5252, which would come out to 269.61hp. Looks like it'd be a nice running car
HTH
--Kyle

and thats at the rear wheels, assuming a 20% loss through the drievetrain that's about 340hp or a bit more at the flywheel.

It was my understanding that the point where the HP curve and the TQ curve cross is point of "lockup" (an almost one to one rotation between the engine and the transmission) in a torque convertor which means less slippage and better efficiency. If it happens later in the rpm run you lose torq at the lower end from slippage. The goal is to get to that point as quickly as possibe in the lower rpm. Mine happened at 4k which means I need a better torque convertor. My current stall is about 2500-2800 which is good on the street but bad at the track. The torque convertor I'm looking at will have a 3400-3800 rpm which will suck on the street but give me that needed kick in the pants off the line at the track.

Huey, you lost me in your explanation of matching a convertor (which isn't hard to do), but the lines always cross around 5200 due to the mathematical relationship betwee the two.

gsxrken said:

Huey, you lost me in your explanation of matching a convertor (which isn't hard to do), but the lines always cross around 5200 due to the mathematical relationship betwee the two.

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I am repeating what the dyno guy told me when he circled the intersection point. Maybe the graph was not calibrated so it was off some? When I say "lock-up" I'm using it as an generic term since a "non-lock" up converter is what we are talking about and it does not really "lock-up" to 100% efficiency. Convertor stuff is very confusing for me as well.

Tim

to pick a converter, just go with the stall speed nearest to 800-1000 rpm lower than your peak torque rpm. It has nothing to do with where the dyno graph crosses, they will alwyas cross at 5252 rpm when shown on an equal scale. If your graph does not cross at 5252, that means the operator graphed the readings on unequal scales.

Clark's car, for instance, would work best with a about a 3300-3500 stall converter.

Clark said:

Hello and thanks to all that helped give me some pointers before I went to the dyno today. I have the results and will need some help with some related questions.

The max torque was 328.63@4300 RPM and from one of the guys that was there at the pull he said to get an estimated HP just divide the torque by .8 and that gives me about 410 HP at the motor. Does this sound right?

The other question was the guy running the dyno said That it could have been better but my torque converter locked up? I could use some help on explaining that.

The great thing was the motor did not break, sounded really good and my car won the best of show in the 69-70 class!

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I disagree. I think you can reasonably divide your rwhp by around .8 to get your hp at the fly with a C6 tranny. I don't know about doing the same for torque. An observation I have is that I am surprised your 393 stroker dynoed at only 328 rw tq. My 362W dynoed at 415 ft lbs of torque at the rw and 325 rwhp. It is a bandied about assumption that a C6 eats 22% of your hp so I divided my 325 rwhp by .78 to get 416 hp and close to 500 ft lbs torque at the fly. My dyno graph printed out both torque and hp at the wheels.

post your graph if you can.

the graph really shouldn't intersect anywhere other than 5200...

This thread is confusing me.

What kind of dynamometer was Dallas Mustang using? The reason I ask is because the Dynojet actually measures horsepower, then calculates backwards to graph torque, using the formula

torque = hp * 5252/rpm

The Mustang brand chassis dynamometer, otoh, measures torque and calculates to hp, using the formula

hp = torque * rpm/5252.

Engine dynos are the same, measuring torque and calculating hp.

So if Dallas Mustang was using a Dynojet, it makes no sense for them to have reported torque and to have no idea about hp.

Confusion number two is about this 4th gear stuff and lockup stuff. Dynojets should be run in 3rd gear, or whatever gear is closest to 1:1.

A lockup torque convertor will not lock up when under acceleration, and it will only lock up in top (4th) gear. So again it makes no sense to talk about lock up in connection with a dyno run. It seems like people are confusing stall speed with lockup. Stall, or flash, is the point at which the driven turbine in the convertor begins to move in response to the driving turbine's motion. The same TC will have varying stall speeds according to vehicle weight, final drive, traction, and engine torque. A really high stall TC makes it hard to use a Dynojet, because the driving turbine is moving so much faster than the driven turbine through most of the engine's rpm range -- meaning that the wheels aren't spinning the drum, and it's the acceleration of the drum vs. E.T. that produces the hp numbers with the Dynojet.

Third, this "divide torque by .8" to get hp has no basis in fact. It will coincidentally give a correct result at about 6500 rpm, because 6500 rpm is equal to 5252 divided by 0.8. I think this poster is confusing the rwhp to fwhp conversion formula. Assuming a 20% drivetrain loss, you would divide rearwheel hp by 80% (0.80) to get flywheel hp.

Fourth, if your hp and torque curves are graphed with rpm as the "x" axis, if the hp and torque curves do intersect, it will ALWAYS be at 5252 rpm. That's because at 5252 rpm, the conversion factor of rpm/5252, or 5252/rpm, will equal "1," and 1 times hp equals 1 times torque.

Fifth, a gas engine's peak hp will almost never occur at the same rpm as its peak torque. So even if you used the correct conversion formula, hp = torque * rpm/5252, plugging your peak torque number into the formula will almost never give you your peak hp. Looking at the conversion formula, and realizing that torque typically falls off steadily as rpms rise, you can see why. As an example, an engine that makes peak torque of 300 lb-ft at 3000 rpm but loses 50 lb-ft by 5500 is making 171 hp at 3000 rpm and 261 hp at 5500 -- 90 more hp on 50 fewer lb-ft. As long as the rise in rpms exceeds the fall-off in torque, hp will rise. Peak hp occurs when the fall-off of torque exceeds the rise in rpm, when the engine's breathing, cam, and compression ratio just can't make any more cylinder pressure. This typically occurs about 1500 to 2500 rpm north of the torque peak. So it's crazy to talk about calculating peak hp from the peak torque number.

Now what was the original poster's question again?

OK, well it is quite possible that the hp and the torque numbers do not intersect, in the case of a low-revving torque motor. That's because the torque starts falling off quickly before the engine can rev to the crossover 1:1 number of 5252 rpm. In the case of such an engine, the hp number will never exceed the torque number. Remember that hp and lb-ft are merely units of measure, and have no objective meaning other than to quantify physical phenomena for purposes of comparing one engine to another.

Otoh, 4300 is a mighty low rpm for peak hp in a gas engine.

I agree with your other information, that a Dynojet or other chassis dynamometer should be run with the tranny in a 1:1 gear ratio. I don't know a lot about the operation of chassis dynamometer, but I would expect that a downshift from OD to 3rd, in the middle of the pull, would sure complicate things. You only have to realize that gear reductions will multiply torque -- the larger radius of the driven gear wheel is like using a longer lever -- and gear ratio increases will decrease torque, to realize that a downshift in the middle of a run creates an apples and oranges deal.

Otoh, if the computer merely started over recording data in the lower gear, and threw out the data from the higher gear, then that part of the run, standing alone, would have meaning. The problem with that would be that the downshift would put the engine at a much higher rpm, so data from the lower rpm ranges would be absent from the final printout. So I don't know what to tell you about this downshifting in the middle of the pull, other than to say if it were me I would want a do-over.

The "divide by .8" suggestion is the correct conversion factor for converting rwhp to fwhp in a vehicle with a 20% drivetrain loss. It has NOTHING to do with converting torque to hp, except that 1/0.80 = 6565/5252, so dividing torque by .8 would work at 6565 rpm, but nowhere else.

5.0ina66 said:

HP= Tq(RPM)/5252, which would come out to 269.61hp. Looks like it'd be a nice running car
HTH
--Kyle

Click to expand...

His horsepower will be higher than 269

That calculation only tells you what hp the car was putting out at 4300RPM. Redo that calculation at a higher RPM and you'll most likely find that the HP goes up..

The problem is, the guys at Dallas Mustang are telling him that peak hp DID occur at 4300.

I still say 4300 rpm is too low for a hopped up gas engine to be hitting peak hp.

I also still say it makes no sense for anyone to say that, to get hp numbers from a Dynojet session -- if that's what it was -- you take the torque results and calculate hp (much less that the correct formula is hp = torque/0.80). With a Dynojet, the calculation is exactly the opposite. The computer measures how much time (call it "delta t") it takes the car to accelerate the known mass of the rollers to various rpm (call it "delta v"). That data gives you an hp number at each delta v, which you then calculate to torque, t = hp * 5252/rpm.

So several things about this make no sense.

180 Out said:

The problem is, the guys at Dallas Mustang are telling him that peak hp DID occur at 4300.

I still say 4300 rpm is too low for a hopped up gas engine to be hitting peak hp.

I also still say it makes no sense for anyone to say that, to get hp numbers from a Dynojet session -- if that's what it was -- you take the torque results and calculate hp (much less that the correct formula is hp = torque/0.80). With a Dynojet, the calculation is exactly the opposite. The computer measures how much time (call it "delta t") it takes the car to accelerate the known mass of the rollers to various rpm (call it "delta v"). That data gives you an hp number at each delta v, which you then calculate to torque, t = hp * 5252/rpm.

So several things about this make no sense.

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The way I read his initial post.. he was told the max torque was at 4300RPM and he was also told to divide his max torque by .8 to get his HP. I don't believe he was told his max HP and max torque both happened at 4300RPM.

And as I understand it, torque is what is measured and HP is what is calculated. I'm pretty sure the only way to calculate HP is by 1st measuring torque. HP isn't something that can be directly measured.

70_Nitrous_Eater said:

The way I read his initial post.. he was told the max torque was at 4300RPM and he was also told to divide his max torque by .8 to get his HP. I don't believe he was told his max HP and max torque both happened at 4300RPM.

And as I understand it, torque is what is measured and HP is what is calculated. I'm pretty sure the only way to calculate HP is by 1st measuring torque. HP isn't something that can be directly measured.

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I totally agree with you. Everything I've ever read says horsepower is calculated from torque. However, I do think what 180out is saying about torque vs. horsepower makes intuitive sense; I can't recall ever seeing a high performance gas burner make peak horsepower at RPM's equal or lesser than that at which it made peak torque. Peak horsepower is normally achieved at a higher RPM range than peak torque.

65up2d8 said:

I totally agree with you. Everything I've ever read says horsepower is calculated from torque. However, I do think what 180out is saying about torque vs. horsepower makes intuitive sense; I can't recall ever seeing a high performance gas burner make peak horsepower at RPM's equal or lesser than that at which it made peak torque. Peak horsepower is normally achieved at a higher RPM range than peak torque.

Click to expand...

That's just it... I don't think he was told his peak HP occured at 4300 RPM, I just think that's how he understood it.

The only way to know for sure is to take a look at the dyno numbers (torque graph) and do the calculations. But without the actual numbers we're kinda stuck here.

so basically, the asshats at the dyno didn't know what they were doing, thus providing the customer with completely useless data.

Clark, if you ever go back to a dyno, go to one that knows what they are doing and can better explain the data/results they give you.

Sorry, but you guys are wrong about the Dynojet. It's an "inertial" dynamometer. It measures how long it takes the car to spool up the known mass of the rollers to a given speed. That's a measurement of horsepower.

By contrast, an engine dyno, using a water brake or an electric motor, measures torque. The brake, aka the absorbtion unit, holds the engine at various rpm points. The dyno measures how much torque the absorbtion unit needs to apply to hold the engine at each one of these rpm points, and pops out a torque curve. The horsepower curve is then calculated from the torque curve.

The problem with using an absorption unit on a Dynojet is tire slip. The Dynojet's steel roller just can't hold 300+ lb-ft of flywheel torque (which gets substantially multiplied by the rear end gear and the tire height) without the tires spinning.

Now having said that, there IS a new model of the Dynojet that does use an electric absorbtion unit to apply some braking force to the roller. But even on this model, it's the measurement of the time it takes to spool up the rollers that is the primary input, and that's a measure of horsepower, not torque. I don't know if the braking force that this new model applies is used to measure torque, or if it's just a way of adding an additional inertial effect to the roller, so that the pull can proceed at a slower pace and/or so that the unit can test bigger engines without increasing the mass of the roller. I do know that a Dynojet's roller simply cannot apply enough braking force to measure the full torque output of a powerful car, because of the wheel spin problem.

As I already mentioned in a previous post, there IS a chassis dynamometer out there that measures torque, it's the Mustang brand. The Mustang uses four rollers, one on the front and rear of each tire. This gives it the ability to brake the tires enough to measure torque, without slippage.

There's another brand of chassis dynamometer whose name I can't remember that also measures torque. This one requires the removal of the driven wheels, and bolts onto the studs. No slippage problem there!