Couple of questions:
If an engine dyno's 530 HP 570 Torque on the stand out of the car what will it have at the crank and wheels?
How much horsepower/torque will a stock rear end handle before braking something or is there no real definite answer? 4 speed by the way.

 

Rule of thumb is a 20% decrease from crank to wheels with auto.....about 15% with standard. So you are going to be putting down about 430-440 at the tire......
The rearend will hold up exactly how good you hook and how hard you beat on it.....there is no real answer for this......

Jebby

 

Thanks for the reply.

 

Your driveline will last longer if the tires you have will 'smoke' instead of 'bite' the pavement. Shock loading with the torque your engine produces is a killer for stock C3 drivetrains. Wheel 'spin' will prevent such high-shock conditions. If your tires get a good bite, ????

 

And while I'm smoking tires the other guy is leaving me lol. But yea I get what your saying.

 

Well put some drag radials on it, do a burnout and see how long the drive-train lasts

 

here we go again.....its not a ratio or percentage loss of what happens from the crank to the rear tires. It is a set amount. So if you knew what your engine HP happened to be and what it made at the rear....Lets say for ease of math 350hp at the crank and 250 at the rear tires it would cost you 100hp to turn your rear wheels.....If you upped it up to 500 hp at the crank you would have 400hp at the rear wheels.

So its all just a guess and an old wives tail to perpetate the percentage guess. If someone had those number and would post what their m20 or m21 with a certain rear gear ratio eats up in horsepower, then we could post that and it would be a better number to repeat and subtract than a given percentage....A library of drive trains would help

If we through out 20% for the ease of math...that 350hp motor would be 70 hp loss. but when we increase the hp to 500hp that same drive train would eat up 100hp. It just doesnt make any sense in the physics world or why we would still be saying it. Its like saying a 75hp volkswagon loses only 15 hp in drivetrain but throw a 450 hp Porsche motor in front of it and it suddenly cant spin the tires with out losing 90 hp?

 

...and post a video!

 

As far as I know, the early C3's that raced with the L88 engines, had the stock Corvette IRS rear ends. The L88's in racing tune were producing about 550+ hp at the flywheel....or so legends state. But these cars were race course driven with numerically low differential ratios...3.08:1(?) and certainly not drag radials. Maybe someone could state the HP that will fail a stock C3 IRS at the drag strip.

 

I've been told that the muncie would hold up to 500 ft/lbs of torque. So I would consider torque valves and where you are with that.....and just stay away from sticky tires. Torque is always higher at launch than HP is so look there first, unless you run a huge stall convertor and want your launch into the HP power band

 

In simple terms, Gearing efficiency (%) = Output power / input power

 

are trying to use that as a percentage for what a motor will put to the tires....it doesnt apply as its the end result of the inout divided by the output. THat wwould mean you would be changing your percentage as you change the engine power

 

There is a output percentage difference in drivetrains. Its called friction.
Even the TH350 has a different friction output percentage than the optional TH400. Why? Because of friction inside the casings.

So, two identical Vettes with exact same 350 CI, same flywheels with exact same H.P. at the flywheel.
But one has the TH350, one has the TH400.
The RWHP will not match

 

I didnt see anyone state the th400 and th350 both have the same amount of parasitic drag... What I did read was that once the amount of hp consumed by the drivetrain in a particular car is established, it is a constant. Your formula it seems would change that amount every time the hp of the engine is changed making the hp loss a variable instead of a constant in that same vehicle unless im missing something.
TH400s waste more power than most transmissions Ive read.

 

right, friction as function of applied force (torque).
To get an idea of the physics of that correlation, rub your inner hand sides against each other with low pressure and 1 rub / sec. (Its like turning your trans by hand)
Now increase force between hands to maximum, stay at 1 rub / sec. (Thats like you engine is turning your trans)
Feel how hands get warm --> frictional power dissipation.
The heat you feel is the loss of transmitted force (= increased drivetrain loss for engines with higher torque compared to engines with lower torque)

This applies for gear teeth too.
Increases as torque increases.. and naturally, as rpm increases.

To simplify it:
Power = Torque x speed(rpm).

Neither a set percentage nor an absolut amount of torque/power are a close enough representation of the happening physics.

Its more like a set amount WITH a substraction or addition of a percentage when comparing two engines.

Engine A: 350 LbFt & HP - peak at 6000
Engine B: 500 LbFt & HP - peak at 6000
Same drivetrain.

Just an example of how i mean it:
Engine A turning a given drivetrain with 100hp loss MINUS 20% (due to lower torque/force to friction losses compared to engine B). So result is like 350-(100*0.8)= 270 RWhp.
Engine B tuning the SAME 100hp drivetrain PLUS 20% loss (due to higher gear teeth force to frictions losses). Result is like 500-(100*1.2)= 380 RWhp.

Thats why low power cars loose so little hp in the drivetrain. Its because they apply so little force to begin with that only very little power dissipates into frictional losses...

Given the identical rpm-windows, a Muncie in a SBC with 350 hp draws less power than the SAME Muncie with a 500 hp BBC.
Once you start applying this behaviour you realize why 1000hp engines can lose 20% with a manual.
And smallblock 200hp engines can loose 18% with an automatic..

 

My response was editted out. You guys are on your own. Good luck

 

"Engine A turning a given drivetrain with 100hp loss MINUS 20% (due to lower torque/force to friction losses compared to engine B). So result is like 350-(100*0.8)= 270 RWhp.
Engine B tuning the SAME 100hp drivetrain PLUS 20% loss (due to higher gear teeth force to frictions losses). Result is like 500-(100*1.2)= 380 RWhp."

I'm not getting this 100 hp loss and then the additional 20%. What is the 100 hp loss if not a drivetrain loss?

 

I understood what you were saying completely…..
It is a “Rule of Thumb” that I mentioned….it is not linear and you are right….it works for lower horse applications and is a rough guideline….

2018 Dodge Scat Pak 392 makes 485 at the crank and about 390 at the tire for about 20%…..BUT! The 2018 Demon makes 840 at the crank and 730-740 at the tire for only 12% of loss…..notice it is the same 100 horsepower…..

Jebby

 

There are additional variables that can affect chassis dyno numbers as well, wheel weight, gearing, etc in addition to the dyno type, Dynojet vs. Dyno Dynamics. Automatics also lose a little more because it's a fluid coupling, not a direct drive like a manual. Typically, a 15-20% drivetrain loss is assumed, but there are too many variables to consider. It's more of a tool to gauge mods made on one car to measure the benefit, and for tuning.

Here's a good article that goes into a little more detail on the subject. Chassis Dyno Guide - The Truth About Chassis Dynos (hotrod.com)

 

Jebby, do you know if the torque at the crank the same for both engines? Was the tranny identical? Was the rear end the same ratio? Exact same bearings and splines throughout both drivetrains? Same lube oils? If not, then there will be differences in efficiency. It can be calculated, gear by gear, bearing by bearing, etc but that's why the rule of thumb is used that is applied across a broad range of drivetrain combos. If one chooses to get into the heavy math, a quick google search on gearbox design and efficiencies is a few keystrokes away.