View Single Post
Old 04-15-2010, 03:28 AM   #39 (permalink)
pgfpro
In Lean Burn Mode
 
pgfpro's Avatar
 
Join Date: Apr 2009
Location: Pacific NW
Posts: 1,552

MisFit Talon - '91 Eagle Talon TSi
Team Turbocharged!
90 day: 63.95 mpg (US)

Warlock - '71 Chevy Camaro

Fe Eclipse - '97 Mitsubishi Eclipse GS
Thanks: 1,315
Thanked 602 Times in 391 Posts
To the OP!!!

HP is not some BS rating like you think it is. Its very real and very needed to compare one vehicle to another. If you understood engine cycle time you would get this.

History:

This is where the constant 5252 comes from.


The word horsepower was introduced by James Watt, the inventor of the steam engine in about 1775. Watt learned that "a strong horse could lift 150 pounds a height of 220 feet in 1 minute." One horsepower is also commonly expressed as 550 pounds one foot in one second or 33,000 pounds one foot in one minute. These are just different ways of saying the same thing. Notice these definitions includes force (pounds), distance (feet), and time, (minute, second). A horse could hold weight in a static position but this would not be considered horsepower, it would be similar to what we call torque. Adding time and distance to a static force (or to torque) results in horsepower. RPM, revolutions (distance) per minute (time), is today's equivalent of time and distance.
Now if we are measuring torque and RPM how can we calculate horsepower? Where does the equation HP=TORQUE * RPM / 5252 come from? We will use Watts observation of one horsepower as 150 pounds, 220 feet in one minute. First we need express 150 pounds of force as foot pounds torque.

Pretend the force of 150 pounds is "applied" tangentially to a one foot radius circle. This would be 150 foot pounds torque.

Next we need to express 220 feet in one minute as RPM.

The circumference of a one foot radius circle is 6.283186 feet. ft. (Pi x diameter 3.141593 * 2 feet)

The distance of 220 feet, divided by 6.283185 feet, gives us a RPM of 35.014.
We are then talking about 150 pounds of force (150 foot pounds torque), 35 RPM, and one horsepower.
Constant (X) = 150 ft.lbs. * 35.014 RPM / 1hp
35.014 * 150 / 1 = 5252.1
5252 is the constant.
So then hp = torque * RPM / 5252

Vehicle Horsepower Rating

In the early 1970s, the Society of
Automotive Engineers (SAE) published the standard J1349 which defined a
standard method of rating an engine’s horsepower. In a few words, the
horsepower rated under J1349 (or SAE net HP) is for a completely installed
engine, including all accessories and standard intake and exhaust systems. In
other words, it measures horsepower at the flywheel and excludes all transmission and driveline losses.
The J1349 standard was not without loopholes. Taking advantage of these
loopholes, some car manufactures are able to inflate their engine horsepower
ratings and therefore rendering the SAE net HP rating less accurate. This led the
SAE to introduce in 2005, new test procedures (J2723) for engine horsepower
and torque (See Certified Power - SAE J1349 Certified Power SAE International).
This testing procedure is optional. Manufactures completing it can be advertised as "SAE-certified".Transmission and Driveline Friction
The power lost in the transmission and in the driveline due to friction and
resistance is where a large portion of power is lost (See PMECH FRIC in Figure 1).
Compared to a manual transmission, an automatic transmission suffers more
losses due to its mechanical complexity and weight (increased friction), the
resistance/slip in the viscous coupling of the torque converter, the transmission
oil pump and hydraulic system. Not surprisingly, a 4 wheel drive vehicle suffers
even more losses compared to a 2 wheel drive vehicle. Also included in the
driveline power loss are the friction from the braking assembly and the friction
from the wheel bearings.
  Reply With Quote
The Following User Says Thank You to pgfpro For This Useful Post:
Christ (04-15-2010)