CYM GIRL

What is the formula?

Mahjik

iTrader: (1)

There really is no formula as each car will most likely have a different drivetrain loss value.

For instance, FD 255hp * 12% drivetrain loss = 30.x difference --> 255hp - 30 = 225rwhp.

So, you would need to know the drivetrain loss estimate to guess from one side to the other.

Hans

I still cant follow the whole % drivetrain loss thing... I ok say i have a transmission and two motors, one makes 100hp and one makes 20hp...be the % logic the 100hp motor will loose less hp through the transmission than the 200...why. wouldnt the transmission never changed the gear rations, the gears mass physicaly never changed, so why would the hp to turn the transmission change. a set hp/number for a given transmission, driveshaft, and differential(axels and all), wheel size/weight, tire weight etc etc etc is the better way to do it. But still you can never get a engine/motor's true hp without from whp


but if you want a formula, its like 15% for manual cars and 20-35 for auto's using most calulators on the web

Evil Aviator

Coefficient of friction.

Hans

Please elaborate Evil Aviator, because if you are telling me that a transmission will rob 15 horsepower from a 100hp motor and 150 from a 1000horsepower motor, I REALLY want to know how you can prove this pyhsically. I have always wondered about this dyno stuff.

I would like to believe there is some magic formula that can give you rwhp from engine hp or vice versa, but there is not. there are too many varibles invovled.

$150FC

when i get home i am digging up a magazine article that itemized "drivetrain loss". if i can find it.

Lith

There are WAAY too many variables to fix a rule onto it. If you want flywheel hp, use an engine dyno Chassis dynos are for tuning, not really for performance comparisons.

Basically, the exact same car can get different power @ wheels readings just from running on different dynos, or even exactly the same dyno with different configurations. Then as other people have mentioned, you have different transmissions loss between front wheel drive, rear wheel drive, four wheel drive, manual, auto, some people argue different diffs/transmissions are more efficient than others etc.

rxseven

Yes it's a percentage not a fixed value for loss thru a gearbox. All gearboxes are rated by their % efficiency. There is always a %loss for a given gear ratio due to energy loss by friction/heat dissipation.

Sponge Bob Square Pants

Usually the higher the HP the car is, the less power that is lost through the drivetrain... or at least that's what I've read, and you know what they say EVERYTHING YOU READ IS TRUE!!

Anyway, I've heard about 15-18% is normal for a sports car, and more like 10-12% for a drag car.

*pertaining to manual transmissions only*

Evil Aviator

Almost every vairiable in automotive performance is a multiplied gain or loss rather than a fixed gain or loss. The automotive magazines and parts manuracturers simply list a fixed gain or loss to make it simple for people to understand. For example, turbochargers multiply boost rather than adding it, cold air intakes multiply air density rather than adding to it, regular air filters multiply pressure drop more than high-flow filters, etc.

Drag and friction can be given as a coefficient. If you look at the various drag and friction equations, you will notice that they are given as a decimal (or percentage) which means that the drag and friction are not fixed values, but rather they change as the force variables change. The main variables are surface area, applied force, and velocity.
http://www.grc.nasa.gov/WWW/K-12/airplane/dragco.html
http://hyperphysics.phy-astr.gsu.edu...frict.html#coe

The type of material and its surface condition affects the coefficient of friction. Once again, notice the decimals (aka percentages).
http://www.carbidedepot.com/formulas...oefficient.htm

As you may have figured out from the above information, a drivetrain's efficiency depends a lot on the materials of its construction, the surface area of its gears, the viscosity of its oil, the efficiency and load of its bearings, the slippage of the gears and clutch, the slop in the driveshaft and gears, the flexing of various parts under applied torque, etc.

bejbis

i motor with 100 hp isnt going to put the same amount of load(pressure) on the drivetrain as a 1000hp motor. more load(pressure) = more friction.

same thing applies to air drag, the faster you go, the more air you displace, the more physical drag your going to have. you can convert the amont of resistance to a percentage, you cant just say, "oh my car is going faster, im still haveing this much drag"

im probally way off base with my thinking. but hey, it seems right to me

-Daniel

CypherNinja

just to throw in my $.02

When running more HP through a gearbox, losses increase because the parts are rubbing together more strongly from increased torque and/or are rubbing together at a greater rate due to increased RPM's.

Also, gear tooth design is a big factor.

KevinK2

you have fixed losses like seal and viscous drag, an tire flex that can get ugly on small rollers. variable load dependent losses include gear mesh friction, and more tire strain losses. my $.02 is to split it in 1/2, based on stock run. Known for fd is 221 dynojet, 255 at eng, or 13.3% loss of 34 hp. 17 hp is fixed losses, other 17 hp is load dependent. This gives load proprtional factor of (221+17)/221=1.077

so run 350 rwhp on dynojet. eng hp = 17 + 1.077X350 = 394 hp.

usual factor is 255/221 or 1.154 for 404 hp ... about 10 hp more.

for 500 rwhp, my method gives 17+1.077x500 = 555 hp (11% increase). old method gives 1.154x500 = 577 hp, a 15% increase with 22 hp more predicted vs my method.

$150FC

couldn't find it. going to europe tonight.
i think it's super street, somebody may have the issue.

crispeed

Well explain this.
200 RWHP with a 10% drvetrain loss equals 220 flywheel hp or 20hp lost to friction.
The same car now with 1000 RWHP with the same exact drivetrain etc. with the above equation would now loose 100 hp to friction? I personaly find that very hard to believe. What about the heat produced by the 100 hp lost due to friction.

Evil Aviator

The "10% drivetrain loss" multiplier is simplified. Some of the loss is based on rpm, some on load, some on temperature, some on inertia, and some on other factors. Therefore, a simplified correction factor such as "10%", "15%", etc., is not very accurate, and it becomes less accurate as the variables change.

Ghetto Physics lab experiment:
1) Rub your hands together. Feel the heat? That is from friction.
2) Rub your hands together faster. Feel the increased heat? That is increased friction from increased velocity.
3) Rub your hands together at the original speed in #1, but press them together with as much pressure as possible. Feel the increased heat vs. step #1? That is increased friction from applied force.

In a nutshell:

Does a given drivetrain have the same percentage of total loss (10%, 15%, etc.) under all conditions? - No

Does a given drivetrain have the same fixed loss (30hp, 100hp, etc.) under all conditions? - No

Are there a lot of variables involved with drivetrain loss? - Yes.

KevinK2

wrong equations.

200rwhp, 10% loss. say stock (fwd likely).

.9xfwhp=200rwhp, so fwhp=200/.9=222, or 22 more. say 11 fixed, and 11 was load related. 11 variable means scaler is 1+11/200 = 1.055 x rwhp.

for 1000rwhp, same car, hp loss = 11 + .055 x 1000 = 66 hp loss, and 1066 fwhp. more reasonable. heat losses can be checked by putting hand on diff or trans after a 20 minute track event.

fstrnyou

who cares what flywheel hp you're making, after all, isn't WHP what's important since that's what is making it to the ground. everyone has drivetrain losses. so get to the bottom line and use wheel horsepower as the measuring tool.

rxseven

Why is that hard to believe? The opposite of this scenario is even harder to believe. If you had a drivetrain that would loose 20 hp with a 200 hp input it would be 90% efficient. If the same drivetrain had a 1000 hp input and would loose only 20 hp, it would be 99% efficient. That would be impossible for the same set of components as it would clearly violate the principles of physics. Frictional losses are directly proportional to the input load.

fstrnyou

i think it's kinda like an alternator. if it takes 10hp to turn an alt. 4000 rpm, what difference does it make how much horsepower the engine has. that's how i look at it. oh well, like i said earlier, go by what the dyno says.

RETed

I really don't understand how people have a trouble understanding the concept of a percentage friction value.

Easy experiment...
Coefficient of friction of an asphalt road surface is constant.
Run your elbow across the surface at approximately one inch per 5 seconds. What do you get? Maybe a tingling sensation.
Now, run your elbow acorss the surface at 60MPH - are you starting to bleed now? Friction is constant, yet the outcome is not identical - if the friction is constant, then you would have equal outcomes no matter when the speed is. Yet, this is not the case. You're ignoring the time/rate factor which is EVERYTHING in friction loss.

Also, be careful with dyno numbers. DynoJets read friction values lower than Mustang/DynaPak/Dyno Dynamics readings, so there is no one set percentage. I find DynoJet's run about 13% - 15%, and Mustang Dyno numbers are 7% to 10% HIGHER (i.e. lower wheel power numbers).


-Ted

1st to 3rd

Good answer. 90% efficiency will always be 90% efficiency.
Power transmitted through a gear box (efficiency) is dictated by the geometry of the gears. There are three forces involved: Tangential force (or power transmitted) this is dependant on geometry, power, and rate. Axial force which is dependant on tangential force and gear geometry. And Radial force which, again, is dependant on tangential force and gear geometry.
Now, the gear box efficiency doesn’t give two ***** about how much power and the rate it’s being applied since the power losses are figured from axial and radial forces. Give it 1hp at 1 rpm and the formulas dictate there will be a certain amount of loss due to axial and radial forces. These forces, geometrically speaking , say there will be a certain amount of loss through friction caused by gear mesh and load transferred to the bearings. Put in 1000hp at 10,000rpm and the tangential force increases, but the geometry stays the same (provided the gearbox doesn’t explode ). Meaning there is more power being transmitted, but the axial and radial loads increase proportionately. Hence, the efficiency of the gear box stays the same.

crispeed

My point exactly!

crispeed

I didn't explain myself properly because I'm in agreement with you but what I wanted an explaination for was the heat generated by friction lost. Wouldn't 100 hp lost to drivetrain friction generate a lot of heat? Were does all this heat go?

fstrnyou

certainly you'd need a trannyfluid cooler by now. and a good sized one sounds like. i'm not saying i'm right or you're wrong, my brain is having a hard time with it.