I have a 4 or 5 part question pertaining to rear ratios.

1. If I changed the ratio from 3.42 to 3.73 (or higher) what would it look like on a dyno sheet?
2. If I am not "Tracking" the car and I just want "leaving the light" quickness - is a basic set-up enough without paying for over-kill ?(now don't jump on me about stories of woe and I know some of you cannot resist saying "get the best"). I am not one to drop the clutch at 3500 at the light and I don't beat the raisons out of the car.
3. Will the dealer be able to tell if this mod was done?
4. At what point do I need to reprogram and if so what is it that needs reprograming and how?

I assume that gas milage would be uneffected until I get on the highway.

 

Good question on the Dyno results. Most dyno instructions just say to run in the gear closest to 1 : 1, they don't say anything about final gearing (or tire size that affects final ratio).

 

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A dyno tune should be done with any major change.

MPG will go down slightly, not by much.

Good Luck

 

AU N EGL,

How do you get HP loss by a rear gear change? And holding 3500 rpm - clutch engaged and fast (but not intantly) apply full throttle works better and less tire spin when the track is sticky with MN6. I have tryed it several ways besides which I spoke with GM's guy, and two drivers from R &T and MT at Virginia International Raceway last summer and that's exactly how they ran all the MN6 C6s.

 

Rember dynos are all run in 4th gear so it is a one-to-one ratio. Durning the run the dyno starts out at 12-1500 rpms and then acceralates slowly to redline, 6000 or 6500 then repeated and corrected for temp and humidity ect.

When you change or lower the rear gear from 3.42 to 3.73 or a 4.10 or putting on larger rear wheels like 19" the power out put changes as well. But dyno numbers dont win races and should only be used as a guide or bench racing over a few

Yes any time you have tire spin you are loosing power.

Most Corvettes have peak torque at 4800-5000 rpms. Some tuners can crate a flat tq curnve from 2500 up to 4800 with is great for getting out of corners and UP to speed. The tq and hp line corss at 5252 rmps with peak HP in the 5000-5500 range.

I also drive a lot on VIR and log about 2000 miles this year on VIR so far. I use Good Year R480 - or R470 Slicks on my C5 coupe, MN6, 3.42 and a 6750 redline. ( other mods too ).

I have driven and tried mutlpe gear and rear end combnations. the 3.73 gear tends to be the most popluar for road racing. I prefer the 3.42 on a MN6 over a MN 12. The worst combination was a C5 ZO6, MN12 with a 4.10. SHifting all the time. Rember the vette has a huge power band and is not limited to a few hundred RMPS so a longer gear is OK to make use of 2000 rmp power band.

 

The dyno measures the TQ on a large massive rolling drum, and makes assumptions about how much engine+tranny+drive line rotational inertia is involved. These assumptions are reasonably accurate when using standard gearing. When you put in a 'bigger' rear end ratio, more of the engine+tranny+drive line rotational inertia is 'visible' to the dyno than its assumptions about same. Therefore, the dyno 'sees' less TQ because the TQ that used to rotate the rear wheels has been absorbed by accelerting the engine+tranny+drive line inertia and thereby does not show up at the rear wheels (as seen by the dyno).

You, on the other hand, feel more thrust at the rear wheels because of the bigger gearing multiplier.

So, the dyno reads less, but you feel more!

Enjoy!

 

Wow, great info! It makes sense in under this context. Now I just need to find out how my unmodified C6 is above average even though I have the 3.42 MN6.

Thanks guys - your great!

 

Hold on... I'm a little confused now. With a 3.72 or higher gear, shouldn't you see more torque on the dyno? When transfering motion through the drivetrain at a given power level (meaning torque, and rpm at the flywheel) the power output will always equal the power input minus frictional losses. Changing the gearing will result in a higher torque and a lower rpm output than before, but equal power (aside from slightly different amounts of friction). This I remember from engineering classes.
Now for real world scenarios, what causes the hp loss on a dyno? Is there really a torque loss on the dyno? If so, why? There's obviously a torque gain at the wheels, which is why you accelerate faster.

I understand the explaination by MitchAlsup, but your engine/drivetrain is accelerated in both situations by the same amount at a given engine speed, it just happens at a different time. Yes, the curves may shift, but shouldn't you still see the same power at 5000rpm?

 

One thing I don't understand is how the rotational
inertia of the drivetrain changes with a rear gear
change (assuming the ring and pinion are the same
diameter as before - which might not be true - I don't
know).

It's easy to see how putting on larger wheels and
tires hurts rear wheel power and acceleration but
unless I am forgetting some physics (always possible)
you'd have the same moment of inertia regardless of
gearing, if all the gears remained the same diameter
and weight. If the ring gear grew in diameter or its
weight, then it's easy to understand how the moment
of inertia changes and how power will be affected.

Pat

Edit: Thinking about this some more, if (and it's a big
"if") the ring gear doesn't change appreciably in
diameter or weight, I don't believe there is a hp change
at the rear wheels. I think you're just seeing dyno
error or variation.

 

I totally agree with you catpat. If the engine is revved from 1500 to 6500rpm in 4th gear, the moment of inertia will actually be lower. A couple extra/fewer teeth on the ring and pinion will be a negligible difference, but when going to a 3.9 gear, the diff. output shafts will spin slower than with the 3.42 so there is actually less inertia to overcome in order to reach 6500rpm.

 

It would seem kinda silly for dyno manufacturers to make such assumptions wouldn't it? Especially since "standard" gearing varies so greatly from one car to the next. Granted I know zilch about dynos, but it would seem really stupid to hard code variables into a dyno given any specific car. Shouldn't you be able to program into the dyno your rear end ratio etc etc if indeed changing your rear end would affect the dyno results?

Anyone own a dyno can comment on this?

 

that sounded wrong to me too. The shorter gearsets are slightly less efficient, so that might be what causes a loss.

The dyno has the gear ratios programmed in, and while you'd get false numbers if you didn't tell the operator you had changed it, it should be easy enough to override/enter manually. The dyno measures power at the wheel, but uses the ratios to calculate and display numbers for the engine after drivetrain losses.

 

If the dyno has the capability of holding the speed constant and measuring the TQ, then the driveline is not being accelerated and you will see the same TQ independent of gearing. However, if the engine is accelerating, with higher gears it is accelerating faster and the dyno does not see the force used to accelerate the engine crankshaft and the tranny,...

At constant rotational velocity (RPMs) this is true. But consider 2 cars identical in every way but one, one with an LS2 one with a 'varient' crankshaft made out of tungsten (which is twice as heavy as steel). Which car will be faster? Answer: the one with the lower rotational inertia! Even when the engine produces identical power at the pistons!

Not only does the TQ at the rear wheels accelerate the car linearly as a whole, but some TQ is absorbed accelerating the rotational elements between the pistons and the rear wheels. This TQ never shows up to accelerate the car, but does show up on a constant RPM (engine) dyno.

lower RPMs for the rear tires, higher RPMs for the engine, and it is this higher RPMs at the engine that are absorbing the TQ the dyno does not see.

Having to accelerate the rotating parts both linearly (like the rest of the dead weight of the car) and in 'rotation" is what consumes this TQ. Rotating parts dont just accelerate with the car, the also accelerate in a rotational sense (RPMs increase). Since this rotational sense has mass, it takes TQ to accelerate it and thereby the force used to accelerate the rotating masses is not available to apply linear acceleration to the car as a whole.

Putting on light weight wheels and tires also increases linear acceleration for exactly the same reasons less rotational inertia to overcome.

The engine driveline is accelerated faster with higher numerical rear gears because you get to redline while traveling at a lower velocity after taking less time to get there! Perhaps an illustrative example is in order.

So, if first gear takes (lets just say) 3 seconds with stock gears and you shift at 50 MPH with stock gearing (3.42). Then you accelerated the engine from 1,000 RPMs to 6,500 RPMs in 3 seconds. If your acceleration was linear across this interval you passed 40 MPH at 2.4 seconds. A 0.5 second shift and 0.8 seconds from 50-60 puts us right at the 4.3 second 0-60 times which seems "about" right.

With agressive rear gears (say 4.11) you accelerate to 40 MPH in 2.0 seconds so the engine accelerates from 1,000 RPMs to 6,500 RPMs in 2.0 seconds, so the acceleration of the engine is 50% faster than the previous case while the acceleration of the car as a whole is 20% faster 0-40. Now a 0.5 second shift and 1.5 seconds from 40-60 puts us at 4.0 seconds 0-60.

So higher gearing gives us 0-60 in 4.0 compared to 4.3, saving 0.3 seconds. Remember, this is an illustrative example.

If you measure the engine on any dyno with the ability to hold a constant speed (no acceleration) you will see the same engine TQ numbers independent of the rear gearing. But when the measurement device is accelerating the drive line inertia consumes more forces with higher gearing.

So, when the engine is at 5000 RPMs and NOT accelerating the dyno will read the force twisting the crankshaft, but when the engine IS accelerating some of the previously measured TQ will vanish since it is being used to accleerate the crankshaft (and other rotating parts) in rotational velocity (RPMs) and the dyno can no longer see some of the twisting force. The higher the force multiplication through the driveline, the higher the loss as seen by the dyno--in all cases the engine is making the same power in the cylinder heads! just not as much of it is showing up at the rear wheel contact patches.

 

And this is the crux of my own personal confusion. Let me
try to ask my question differently. Please correct me where
I've made a mistake.

I assume we're talking about an inertial dyno. An inertial
dyno produces HP numbers by knowing the moment of
inertia of the drums, and the drum's rpm at a bunch of
points in time..

From these values, it can determine the acceleration of the
drum. It then calculates HP through a relation like:

HP = (moment of inertia * acceleration * drum circum) /
(time (measurement interval)* 550 (conversion factor) * rear end ratio)

Assuming for the moment that changing the rear end
ratio from 3.42 to 3.90 or so doesn't significantly alter
the moment of inertia on the ring gear or any other part
of the driveline, then only two factors change in the HP
calculation with a rear gear change:
1. rear end ratio
2. acceleration

In this case, when the rear end ratio goes up, say from
3.42 to 3.90, the acceleration will increase by a
proportional amount (14%). This means the HP number
will not change because both numerator and denominator
are changing by proportional amounts, because nothing
else (including the moment of inertia of the entire driveline)
has changed.

So if the moment of inertia of the driveline doesn't change
with altered rear gearing, I still cannot see how the dyno
can show a HP loss with a steeper rear gear.

Can someone explain using the relation (or a corrected
one) I described above?

Pat

 

I get it now, and Pat, I think this is what you're not seeing too. When you run a 4.1 rear on the dyno, the drum has a higher rate of acceleration than a 3.42 which is due to more torque being applied to the drum. The rear end ratio is entered into the dyno and factored into it's calculation so that the output is what you would see at the wheels if there were a true 1:1 ratio (I would think) between engine rpm and the wheels. So you would think they should be the same.

However, with the 4.1 gears the engine revs up quicker. The engine does work in accelerating the internals, and must do so at a higher rate. Since power is defined as the rate at which work is done, it takes more power to accelerate the engine/drivetrain internals. With more power being used to do this, there is less at the wheels.

 

Let me at lest say, you are very precise about what you don't know....

The moment of inertia of the driveline components HAS not changed,but the moment of inertia of the driveline "visible at the rear wheel contact patches" HAS changed! And it changed because the driveline is being accelerated (rotationally) faster with the new rear gears. The driveline rotates faster and the driveline accelerates in rotation faster.

In fact, the MoI at the rear wheel contact patch increases with the square of the gear multiplication (assuming the tires aren't spinning)! Since the driveline is spinning faster (by the new/old ratio) and since the driveline is spinning faster there is more force at the contact patch, so the car is accelerating faster and, thus, the engine is accelerating quadradically faster!

 

I think you mean that the rate of angular acceleration changes not the MOI. After all MOI is the sum of the rotating masses multiplied by the square of the distance it acts from the axis. Unless you've got gear teeth flying off as it rotates that value is a constant. But you're definately right. This is a conservation of energy problem differentiated over time. Energy, or work, measured in joules, doesn't change. Flywheel - drivetrain = wheels
Thus, torque measurements shouldn't vary. But the question is power, which is work differentiated over time. This is Joules/second aka Watts, or with a conversion factor Hp to us Americans. Flywheel power is constant, drivetrain losses increase, and less power gets to the wheels.

Wow... I feel like I'm back in college again

 

Homework and POP quizes are for life.

 

You might be right here, which is why I tried to use the MoI as seen at the rear wheels. The MoIs of the components remains constant (baring the tungsten crankshaft sub-diatribe). But the drive line components are operating at higher RPMs, and thus to accelerate a given percentage, the engine RPMs pass through more RPMs than before.

TQ measurements without acceleration do not vary since MoI stuff is getting multiplied by zero (the acceleration going on). While TQ measurements with accelerating have all that MoI stuff going on, and subtracted out because this force/power does not go to accelerate the car (forward) but to accelerate the rotating pieces (round and round).

Power is energy differentiated over time, Work is force through a distance. subtle difference. But you are on the right track differentiating from energy towards force or integrating force up through energy. Power is mearly the first integral of force versus time and the first differential of energy with respect to time.

And you thought finals were over!

Monty Python: Nobody expects the Spanish Inquisition......

 

Well I'm glad we straightened everything out. I hope everyone else remembered to take notes.