Dyno result question?????

Newport Viper · Jun 27, 2002

If you have 3.55 gears will it be accurate?

Performin Norman · Jun 27, 2002

If I remember correctly, I think you gain torque but lose horsepower. Someone else may step in and correct me if I am wrong.

FUSCUCLA1 · Jun 27, 2002

What were your results?

99 R/T 10 · Jun 27, 2002

I was told the computer takes into account all the variables to include the gears and should produce the numbers accurately. That's what I was "told", don't know for sure.

Mike

Newport Viper · Jun 27, 2002

Allen,
Didn't do it yet.
Does anyone know for sure? Thanks for you replys so far.

Mark Young · Jun 27, 2002

Do a search, we covered this topic a while back. You will NOT get the same results with a 3.55 than with a 3.07. Its basically the same principle as why you would not expect to get the same results if you dyno'd in 3rd gear or 5th gear rather than 4th gear. I would expect it to be "close", percentage wise, but the curves will certainly be different, as you would expect.

Since there's no place to specify the gearing on the dyno computer, there's no way it can "magically adjust" itself.
I don't know why someone who has done a gear change hasn't done a before/after dyno run???

Joseph Houss · Jun 27, 2002

I was under the impression that horsepower is horsepower, whether with a 3:55, 3:07, Viper, Mustang or Vette. There might be slight differentials based on wheelspin issues with a shorter gear, but the results should always be close. I don't believe you actually "dial in" a specific brand or year of car on a dyno... just hook it up, and start "the pull".

Can anyone confirm?

joe117 · Jun 27, 2002

Torque is real and, of course, horsepower is real too. Torque in itself means little. A coiled spring produces torque. It will sit there till it rusts away, applying torque the whole time. Horsepower is a measurement of torque at a given RPM. The results of this, torque vs time measurement, go into a simple formula to give horsepower. A torque measurement taken this way includes time and will give you a power measurement. Horsepower is very real. Please, Maxwedge, tell us why horsepower isn't real.

Venom Lover · Jun 27, 2002

So all the hypothesizing is interesting, but I can tell you from personal experience: rear end ratio will affect the results as measured by a Dynojet. I did a before and after dyno when I put the 3.55's in my GTS, and I went from 455 rwhp to 445 rwhp (SAE corrected, peak).

Is this a real effect? Of course not. Is this what happens on a Dynojet? Apparently. Do I know why? No.

Ulysses · Jun 27, 2002

I take it all back. You make a convincing argument.

joe117 · Jun 27, 2002

Maxwedge,
That's a very good piece. It's well written and the science is right on. However, you wrote,
Horsepower doesn't exist anyway..Toque DOES!!!
These are your words. You won't find that statement in the piece you posted. I wonder if you read the piece you posted. Read the part about the high torque windmill.
We all know that horsepower is derived from a torque at rpm measurement. Just because horsepower is derived partly from torque doesn't mean that horsepower doesn't exist. You said "horsepower doesn't exist" I thought that's was what you meant. If that is what you meant then you are wrong. If you don't agree with me then please, in your own words, explain your statement, not some other position, your statement.

USAF BAD ASP · Jun 27, 2002

<BLOCKQUOTE>quote:<HR>Originally posted by Maxwedge:

Horsepower doesn't exist anyway..

<HR></BLOCKQUOTE>

<BLOCKQUOTE>quote:<HR>Originally posted by Maxwedge:

again I agree HP exists

<HR></BLOCKQUOTE>

the maxipad ***** strikes again!!!!

Newport Viper · Jun 27, 2002

Venom Lover,
Can you post any graphs? I just can't figure out why the difference? What about the torque!

bitya · Jun 28, 2002

I have done plenty of 2-3-4-5-6-gear pulls on many diff vehicles to satisfy my own curiosity on the results. Most of the time there has been a 1-2 hp diff, when there is more its due to other variables. Most common is being way over cammed, dropping down a gear shows a lot better hp,it gives the engine the abilty to acclerate better. In your case the engine is still accelerating the same rate but in reality dyno rollers are slower cause of the gear change, get into 5&6 different story now the gear change will allow the engine to acclerate quicker witch will show up as a better accleration time. The more time i spend on the dyno, the more i realilize the endless combinations that exist.

Newport Viper · Jun 28, 2002

bitya,
So in you opinion a 3.07 and a 3.55 is 1-2 hp. Right?

treynor · Jun 28, 2002

One plausible explanation for the small but consistent difference in dyno readings between 3.07 and 3.55 rear gears is that the rear diff absorbs more power to friction with 3.55 gears than it does with 3.07. The reasoning's simple enough -- for a given revolution of the rear wheels (distance, as measured by the dyno) there are more gear mesh events in the diff, and thus more total distance traveled by the meshing gears, thus more friction, etc. etc. If we assume the Viper loses 12% of its power on the way from crankshaft to dyno, the diff accounts for perhaps 5% of that loss. Going from 3.07 to 3.55 gears would add an additonal 5% * (3.55 / 3.07) loss to friction, or about 3.5 HP worth...

Ulysses · Jun 28, 2002

A dynamometer measures torque, then does a calculation to find horsepower. Torque is torque no matter what gearing you may have in the rear diff. Dyno results should be pretty close if not the same. How a car performs off the dyno with different gears will change.

joe117 · Jun 28, 2002

Maxwedge,
When you say,
My Beef is the fact that they expect me to believe that EVERY engine has more HP than torque automatically at 5252RPMs
It shows that you still didn't read and understand the piece you posted. The fact is that every graph showing horsepower and torque piloted on the Y axis against RPM plotted on the X axis, will always show horsepower and torque having the SAME number @5252 RPM. For example, an engine might show 290 HP @5252. This engine will also have 290 ft/lb of torque @5252. The crossover is an artifact of the formula used to derive horsepower from the torque at a given RPM. The fact that they always have the same number @5252 doesn't mean that this is the peak for torque or horsepower. It simply means that the standard formula was used. If the scale of the graph is such that the torque and horsepower numbers are aligned on the Y axis, the lines of the graph will actually cross @5252. If the scales are not the same then the lines will not cross but the values taken from the scale used for horsepower and torque will still be the same @ 5252 RPM.

Ulysses · Jun 28, 2002

Well.... Max's post on the physics of torque and horsepower peaked my curiosity. Physically it made a lot of sense, but I was still a bit confused. If reduction gearing multiplied up the torque, why can't we just gear it down and get tons of torque and therefore tons of HP? Why doesn't the dyno report this great increase in torque at the rear wheels? Why are people reporting greater HP's with a lower gear?

Being the geeky engineer that I am, I took out the old physics books, blew off the dust did some reading, and then did some research on dynos. It opened my eyes to the light. So without getting too much into the physics...

A dyno does not report rear wheel torque, it reports flywheel torque with some losses, otherwise known as engine torque.

Rear wheel torque (Tw) is related to engine Torque (Te) and is a function of final drive ratio (f) and the selected gear ratio (g):

Tw=F(f,g)*Te

In simpler terms this is just:

Tw=f*g*Te

So for an ideal system of 500ft-lbs engine torque, a 1:1 4th and a 3.07:1 rear, this would lead to 1535 ft-lbs.

But the dyno does not report that, it reports 500ft-lbs if the system is ideal (no loss). How does it do that? It has no idea what tire size you have, what gearing you have.

It turns out, all the dyno cares about is RPMs. Engine RPMs and the RPMs of the dyno drum.

Wheel RPM is directly proportional to drum RPM. The reasoning is that at the point of contact between wheel and drum, they have the same circumferential speed.

vw=rw*ww=vd=rd*wd

vw=the velocity of the wheel
rw=the radius of the wheel
ww=the angular velocity of the wheel
vd=the velocity of the drum
rd=the radius of the drum
wd=the angular velocity of the drum

the angular velocity (w) is just 2*pi*revolutions per second

With this relationship we get

wd=ww/A
where A=rd/rw

Wheel RPM can be related back to engine RPM through F(f,g)

we=F(f,g)*ww

that gives us the relationship

wd=ww/A=we/{A*F(f,g)}

Torque is given by T=F*r when force is applied perpendicularly

F=Force

By Newton's Third Law, the force on the drum by the tire is the same as the force on the tire by the drum

Fw=Tw/rw=Fd=Td/rd

which simplifies to

Td=A*Tw

remember that Tw=F(f,g)*Te

therefore =A*F(f,g)*Te

A*F(f,g) is just the ratio of the RPMs A*F(f,g)=we/wd

so Td=Te*(we/wd)

or

Te=Td*(wd/we)

So if the torque on the drum goes up with a relatively small decrease in the drum RPM via a gear swap from 3.07 to 3.55 you get a larger engine torque reading for a given RPM.

Going back to the equation HP=Torque*RPM/5252 this means that at a given RPM, your HP reading will be higher at the wheels for a 3.55 versus a 3.07.

We also cannot constantly increase the HP because we are limited by the physical size of the differential. To gain enormous HP from gearing we would have to install enormous gears. Sizes beyond the size of the differential design.

At least, that's how I interpret all this to mean.

Newport Viper · Jun 28, 2002

OK, All that are now confused raise you hands!!!
So, after all that being equal how much of a difference between the same car with 3.55/ 3.07 in torque and h.p.on the dyno?
let's say my car reads 385h.p. with 3.07
what would it be with 3.55?

MichaelP · Jun 28, 2002

Gears do not change the HP.

Right?

treynor · Jun 28, 2002

Heh heh heh. Never fear, Treynor's here

<BLOCKQUOTE>quote:<HR> Te=Td*(wd/we)

So if the torque on the drum goes up with a relatively small decrease in the drum RPM via a gear swap from 3.07 to 3.55 you get a larger engine torque reading for a given RPM.

Going back to the equation HP=Torque*RPM/5252 this means that at a given RPM, your HP reading will be higher at the wheels for a 3.55 versus a 3.07.

We also cannot constantly increase the HP because we are limited by the physical size of the differential. To gain enormous HP from gearing we would have to install enormous gears. Sizes beyond the size of the differential design
<HR></BLOCKQUOTE>

Your analysis is correct up to this point. The torque as measured at the dyno drum is equal to the torque as measured at the engine, multiplied by the effective gearing between the engine and drum (which includes wheels) minus frictional losses in the system, or as you put it Te = Td * (Wd/We).

However, the conclusion you then reach is incorrect. First, consider how the dyno calculates engine torque based on observed torque at the dyno. The dyno "knows" Td (from the observed acceleration of the dyno drum and the known rotational intertia of the drum), it knows Wd, and it knows We from the inductive monitor of the spark plug wire. So it calculates engine torque exactly by calculating Te from Td, Wd and We as you describe above.
Now, consider the change from a 3.07 to a 3.55 gearing, an increase in ratio of 15.6%. In this case, at a given engine RPM the torque observed at the dyno will have gone up by 15.6%, and the dyno drum speed will have gone down by 13.5%, a decrease of the same ratio, since (100% + 15.6%) * (100% - 13.5%) = 100%.
So what will the dyno calculate with the new gearing? The new Te' = Td' (which is 15.6% larger) * Wd' (13.6% smaller) / We (constant). But that is of course the same as the original Te, so the dyno isn't "fooled" and will report the same torque (and, by extension, horsepower) regardless of gearing.

Still confused? An example will make everything clear

Consider the ideal car (for this example, anyway) -- the Doug Levin "Dominator" Viper. Measured at the engine, this beast has 1000 ft/lbs of torque @ 5000RPM, and is so well assembled that there is ZERO friction in the driveline. To put all this power to the ground, the owner installs the special Parts Rack "TractionMaster(tm)" 2.00 rear gear set, then takes the car to the dyno for some brag sheets. As it turns out, the dyno drum on his local dyno is exactly 50 times as large as his rear wheels, so 50 revolutions of the rear wheel will result in 1 revolution of the drum. Thus, in the 1:1 4th gear the "Dominator" has a 2:1 drive ratio and a 50:1 wheel/drum ratio, so the dyno measures 50*2 = 100 times the torque being output by the engine and similarly notes that the dyno is rotating at 1/(50*2) or .01 times the speed of the engine. It thus dutifully reports that, at 5000 RPM,

Te = 100,000 ft/lbs (Td) * 50 RPM (Wd) / 5000 RPM (We), or 1000 ft/lbs of torque at the engine, exactly what we expect to see. Similarly, the dyno reports that HP at that RPM is (1000 * 5000 / 5252) or 952 HP.

Our fearless owner, hoping to get even more power, orders the Parts Rack "Ulysses Horsepowerinator" rear gear set with a 20:1 ratio made possible through the use of special titanium cryo-treated ultra-heavy-duty gearsets. These installed, he heads back to the dyno (mostly in 5th, since 1st gear is now only good for 8 MPH) and fires up the beast. Again in 1:1 4th gear, the dyno now finds that the torque at the dyno is 50*20 or 1000 times the torque at the engine, and similarly the dyno is spinning at only 1 / (50*20) or .001 times the speed of the engine. Thus, it reports that:

Te = 1,000,000 ft/lbs (Td) * 5 RPM (Wd) / 5000 RPM (We), or 1000 ft/lbs of torque at the engine, and of course again 952 HP.

Moral: Gears don't make horsepower. Gears multiply torque and divide velocity.

Newport Viper · Jun 29, 2002

Brain now clear!

Ulysses · Jun 29, 2002

Ah!

I over looked the most important relationship:

F(f,g)

which shows up in two instances:

Tw=F(f,g)*Te
and
we=F(f,g)*ww

Tw'/Tw=f'/f=3.55/3.07=1.1564 which is a 15.64% gain and at the same time

ww'/ww=f/f'=3.07/3.55=.8648 which is a 13.52% loss

which, of course, leads to Te' = Td' (15.6% larger) * wd' (13.6% smaller) / we

In trying to understand why some see horsepower gains, I jumped the gun.

So, this leads to the conclusion that the horsepower gains people are seeing are caused by increased efficiency in transfer of power by the new gears, or better environmental conditions from one dyno run to the other, or other external influences.

Dyno result question?????

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