Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations GregLocock on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

Calculating power at the wheels (newbie question)

Status
Not open for further replies.

AlmostClever

Automotive
Jan 25, 2010
5
I'm trying out some ideas for improving drive train efficiencies. For this, I'm trying to model the power necessary at the wheels of a truck.
To validate this, I've entered the specs of a real truck, and the Heavy-vehicle version of the Urban Dynamometer Drive cycle.
My calculation is that the power necessary is dv/dt * the forces to move the truck.
Problem is, I'm getting numbers which are about three times the max output of the engine on the actual truck.

I've considered dividing by pi :), but I would like to know what I'm doing wrong.

Please forgive me if this post is inappropriate. I'm just a big-diesel mechanic toying with some new ideas for hybrid powertrains.
Let me know if I shouldn't have posted here, and I'll ask for the post to be removed.
Thanks

 
Replies continue below

Recommended for you

Simply put.

With no losses, the HP seen at the engine flywheel or output, would be the same at the rear axle or where ever the power is sent.

What does happen is torque multiplication through the gearing. This is why an under powered class 8 truck,(all of them are under powered) can pull 80,000 lbs up a steep grade, though not very fast.

As you can see if you multiply the torque, and reduce the rpms the HP will stay the same no matter what gear its in.
That is with no losses of course.

Its the LBs/HP thats important for moving a vehicle up a hill.

Thats why a 1800 lb Geo metro will beat any loaded class 8 truck, the Geo has 55 hp, and say the truck has 600 hp.
32 lb per HP for the Geo, 133 lbs per HP for the truck.
Lets see that means for the truck to do what an under powered Geo metro will do the engine in the truck needs to be 2500 hp and thats to equal a metro. Pretty bad huh??
 
dicer's elegantly simple computations drive home the importance of driveline efficiency and gear ratio selection for maximizing performance of large trucks. Use of spiral bevel gears at the differentials boosts efficiency to about 95% to 98% from about 80% to 85% for hypoid gears for that portion of the drivetrain. Within the transmissions (main or auxiliary) either spur gears (noisy but strong, efficient, and with no axial loading for bearings) or helical (quieter, nearly as efficient, but with significant axial loadings for the bearings), each contact will have an efficiency of 97% to 98%.

As a truck driver (in a former life, so to speak), I was always impressed by the wide variety in the performance capabilities of different trucks due primarily to the selection of gear ratios available--actual engine power always proved to be a secondary factor. Trucks with transmission ratios that were selected for the purpose of keeping the engine operating within its optimum speed and power range always performed very poorly compared to the trucks that had transmission ratios that were chosen to be progressively closer as road speed increased. At lower speeds, relatively wide steps were not much of an issue because full engine power and torque were rarely needed for acceleration or hill climbing. It was only as road speed increased significantly that using full engine power really mattered very much. At low road speeds, as long as the minimum engine speed was well above idle speed, the available torque was always more than sufficient to accelerate the truck because of the great torque multiplication provided by the "lower" transmission ratios.

The trucks that usually performed the best had transmission ratios selected to provide increments of about 5 mph at maximum normal engine speed for each gear change. When going from near level road to climbing a hill, each drop in gear selection then only lost 5 mph. Usually, 1, 2, or 3 gear changes would be necessary to climb a hill losing 5, 10, or 15 mph in increments before topping the hill.

With transmissions that were selected to keep engine speed within its optimum speed range, the truck could easily experience a drop of 15 to 20 mph in just one increment at higher road speeds when beginning to climb a hill. If a second gear change became necessary, the next increment would result in a drop of an additional 10 to 15 mph for a total loss of 30 to 35 mph. Almost always, after such down-shifts, the road speed was limited by engine rpm red-line and not available power. At low road speeds, the ratio selections were so close together that it was almost always more practical to skip one or two ratios at at time while accelerating in the "lower" gears. (I always guessed that this type of transmission was designed, selected, and purchased by people who never had to drive a truck for a living. I remember these types of transmissions getting lots of positive "press" at the time, and I was never able to make sense of all the "hoopla" for something that performed so very poorly in practice.)

Staying with dicer's nice, clean basic approach and looking very simplistically at a truck with a 600 hp engine, a main transmission, an auxiliary transmission, and a single fixed ratio at the differential, effective available power at different overall ratios can be represented as follows:

In lower ratios where neither transmission is in direct drive:

600 hp at engine
600 hp x .98 x .98 = 576 hp at shaft between main and auxiliary transmissions
576 hp x .98 x .98 = 553 hp at shaft between auxiliary transmission and differential
553 hp x .85 = 470 hp to drive axles with hypoid gearing (130 hp drivetrain losses)
553 hp x .97 = 536 hp to drive axles with spiral bevel gearing (64 hp drivetrain losses)

With one of the transmissions in direct drive:
600 hp x .98 x .98 = 576 hp at shaft between auxiliary transmission and differential
576 hp x .85 = 490 hp to drive axles with hypoid gearing (110 hp drivetrain losses)
576 hp x .97 = 559 hp to drive axles with spiral bevel gearing (41 hp drivetrain losses)

With both transmissions in direct drive ("top gear"):
600 hp x .85 = 510 hp to drive axles with hypoid gearing (90 hp drivetrain losses)
600 hp x .97 = 582 hp to drive axles with spiral bevel gearing (18 hp drivetrain losses)

From this, it is easy to see the distinct efficiency and performance advantages of fewer gear contacts, using the highest efficiency gear contact types, and having the most truly practical selection of gear ratios to serve the actual needs of the truck and the duty for which it is being designed.
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor