posted January 11, 2006 05:00 PM
Edited By: BA on 11 Jan 2006 21:50
Attention Gearheads and Mechanical Engineers!! There are several of you that are either WAY smarter than I am, or you surf the Net way more than I do. I'm looking for some info. Quotable, referencable info.
It is common knowledge that a normal car or 2-wheel drive truck will have anywhere from 10% - 22% loss of horsepower through the drivetrain.
ie. an engines crank horsepower may be 300, but the power at the road will be 10% - 22% less.
I NEED to know how much is lost in the rear differential and cannot find the data. I need tests and results, with referenceable data.
I've searched for hours using google but haven't found a referenceable resource. (billybobs website doesn't count!)
can anyone find a reference for this info?
The closest I've really come is in reading about the "spin-down test" using a Dynojet of all things. I guess it's close, but that's a method, not hard numbers.
posted January 11, 2006 05:05 PM
The drivetrain loss on a motorcycle is consistently 12%. If you take the 185BHP claimed on a zx12r (bikez.com) and subtract 12% It is exactly 162.8 horsepower.. at the wheel..
posted January 11, 2006 06:11 PM
try looking into the heavy equipment, trucking industry side. Of course the numbers will be higher due to weight. That said. I would believe that the hp loss in the dif would be on a sliding scale. The fast you go the more hp power is used. So there is a variable as always. I believe that there is a entry on hp loss in the TMI manual.
____________
everytime man makes something idiot proof, God makes a better idiot............
posted January 11, 2006 06:49 PM
BA,
A 12 bolt Chevy takes less hp than a 9" Ford and the 9" takes less than a Dana.............Does that help? :-)
Sorry I can't help ya.
Dan
posted January 11, 2006 06:53 PM
Edited By: trenace on 11 Jan 2006 19:01
I know I've seen a figure from a decent source for typical loss from a 90 degree change in direction of power flow. I believe it was 5% but I am not positive. I also believe I recall that figure within a writeup on the MotoCyzsyz engine.
I don't know if you consider this a "Billybob" source or not, but here's a figure given for differentials:
As it happens, most people, including those that know what they're talking about, that put up articles on websites don't include references. You may be SOL there.
posted January 11, 2006 07:10 PM
Hey there Dan. Long time.....and, no, that didn't help. lol! (I do have a Ford 9" under my project 69 RS Camaro though) I need hard percentage-range numbers. it. 5%-8% or, 8% - 13%, etc.
Trenace, ya know, I had seen that too. I believe the numbers there, but, need to show more proof to settle a debate.
posted January 11, 2006 07:15 PM
Edited By: trenace on 11 Jan 2006 19:17
It may require a trip to a university bookstore or library and thumbing through some reference works. I also am coming up with nothing with references on the web.
It's possible you might find a value claimed for a specific system in a US patent -- I didn't read through the patents but here's a search turning up some possibilities.
posted January 11, 2006 07:44 PM
Go find a Bosch Automotive Handbook
I'm pretty sure it's in there. I can check when I get home in a couple of weeks if you can't find a copy.
D.
Drivetrain Losses
There are two ways to measure horsepower with dynos: at the flywheel or at the rear wheels. Unless you're an automobile manufacturer or have a very big dyno cell, most people don't test engines in the same configuration they will run in the car. For example, you probably won't have the exact same exhaust system with all the right bends and routing or the complete accessory system with all the belts and pulleys or fan, all of which rob some horsepower. Also, the engine dyno doesn't account for the most important source of horsepower-robbing friction: the rest of the drivetrain.
Drivetrain loss generally ranges from about 15-17 percent for a manual transmission to 20-25 percent for an automatic. In other words, an engine producing 400 hp at the crankshaft may transmit as little as 300 hp to the rear wheels. Automatics eat more horsepower than manuals primarily because they have heavier rotating assemblies and torque converters, which cause some horsepower to be lost through slippage (lockup-style torque converters are more efficient, but most are not recommended for high-horsepower applications). The type of rearend installed in the car can also affect rear-wheel output. For example, a GM 12-bolt or Ford 8.8-inch eats less power than a Ford 9-inch due to the way the ring-and-pinion gears intermesh. The bottom line is that it's how much power you ultimately put to the pavement that counts, and a chassis dyno is the best way to measure that.
Another issue to consider with an automatic is the torque converter. When chassis dyno'ing any engine with a torque converter, the dyno won't start taking accurate torque and horsepower readings until the converter has reached its true stall speed and is transmitting 100 percent of the power to the rear wheels (in reality, most converters never transmit 100 percent, because there is always some loss to hydraulic slippage unless it's a lockup-style converter). The chassis dyno thus becomes a very effective device to determine your converter's specific characteristics as installed in your car, as we learned with our SuperNova.
posted January 13, 2006 04:25 PM
I looked for the Bosch book at 3 bookstores but none of them carried it. They all could order it for me, but that's about it. Heck I can do that on my own!
Doug, if you read this,...could you please check in the book when you return. I'll send ya a PM too since I expect this thread to die.
posted January 14, 2006 11:54 AM
Thanks for that, Larry. That issue has been driving me nuts for months......
Did you get my Email about the bike shoe?
D.
posted January 14, 2006 05:17 PM
i've been out of electronics for a few years so maybe i'm missing something, but the "Problem" and the resulting conclusion appears to be incorrect because of an incorrect assumption, namely that after closing the switch, the voltage on each cap will be half of the initial V1. assuming no loss thru the switch, then it would be
V1 = V2 = V0/rt(2)
Of course there will never be a truly lossless transmission in such a configuration, but regardless the suggestion that V1 = V2 = V0/2 is incorrect (to the best of my current knowledge).
quote:Of course there will never be a truly lossless transmission in such a configuration, but regardless the suggestion that V1 = V2 = V0/2 is incorrect (to the best of my current knowledge).
heheh heheh you said "current" knowledge. heheh heheh
HOME
NEW TOPIC
posted January 11, 2006 05:00 PM
Edited By: 
