## STUFF ABOUT TORQUE, HP, AIR&ROLL RESISTANCE

Discussion about the Hemi in general.

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### STUFF ABOUT TORQUE, HP, AIR&ROLL RESISTANCE

Subject: Lotta stuff
Lotsa Stuff:
If you ever wondered how much HP is required to pull your vehicle along whether it be an 18 wheeler or motorcycle here's a method that is fairly accurate. There are two very small error conditions which tend to offset each other and are not addressed. You must know as close as possible the vehicle weight and decide at what speed in mph you want to test for and mph must be converted to fps(feet/sec) by multiplying mph X 1.46. Example: if testing for 60 mph we calculate the KE(kinetic energy) at 65mph and at 55 mph and then coast down from 65 to 55 mph and use a stopwatch timer to determine the number of seconds it took for the coastdown. From the KE at 65mph we subtract the KE at 55mph then divide by T(coastdown time in sec) and then divide by 550. The final number will be HP required at 60 mph. Now the formula to calculate KE is as follows: KE = W(weight in lbs) X V squared(V is ft/sec) divided by 2G(64.32).
As an example consider a 3000 lb car at 60 mph and a coastdown time of 10 sec from 65 down to 55.
KE at 65 is 3000 X 9006/64.32 = 420,056 ft/lbs
KE at 55 is 3000 X 6448/64.32 = 300,750 ft/lbs
The difference is 119,305 ft lbs divided by 10sec = 11930 ft lbs/sec divided by 550 ft lbs/sec = 21.69 HP.
In testing the coastdown time choose a fairly level road with as little wind as possible and for the above example drive at 70 mph, shift to neutral, start timer as speedometer goes by 65 and then stop timer at 55; you may want to do about 3 and take the average.
One of the small errors is that the flywheel effect of the tires, wheels, brake discs/drums, axles and drive shaft all help to keep the car from coasting down, but do not help to propel it along. If we knew the weight and "radius of gyration" of these and then calculate the power between 55 and 65 mph it could be deducted from the final number. In the above example for 1 wheel/tire/brake weighing 30 lbs and with a ROG of 1 ft the power would amount to only about .07 hp.
It seems that yet another small error is due to calculating the KE at 55 and 65 mph the V is squared and perhaps it would be a little more accurate if we used the "root mean square" of these rather that the common mathematical average. The RMS of 55-65 is 60.2.
So in the first error we would need to deduct a little hp and in the second we add a little to the final number. And these amounts are very small and tend to offset each other.
More Stuff:
Reading magazine articles indicates TORQUE and HORSEPOWER are probably the most misunderstood aspects of an engine. Torque is a twisting force and usually is expressed in FT LBS. Consider a torque wrench with a 1 ft. long handle and assume it's on a bolt head and the handle is in the horizontal position. Assume a 50 lb. weight is suspended 1 ft. out from the nut and we note that the reading is 50 FT LBS.
OK, now lets assume we can turn the wrench a complete turn and then the distance traveled would be 2 Pi X 1ft.= 6.28 ft. OK, now we say 50 X 6.28 = 314 ft lbs of work.
Horsepower is expressed as 33,000 ft lbs of work per minute or 550 ft lbs per second. If we had put about 87.5 lbs force on the torque wrench and made a complete turn in 1 sec. then we would have been working at the "rate" of 1 HP.
Keep in mind that torque is a force and horsepower is only a calculation. It is possible to have torque and zero HP, but you must have torque to develop HP.
There is one and only one RPM where the torque and the HP will be the same numerical number. That number is 5252 RPM. The reason being is when you divide 33,000 by 2Pi you will get 5252. At less than 5252 the torque will always be more than HP. At greater than 5252 the HP will always be more than the torque.
Some Air Cond. Electric motors are marked in ft. lbs. of torque. If a 1750 rpm motor is marked for 60 ft lbs then you know it's 20HP. 1750/5252 X 60 = 20.
BMEP=TORQUE X 150/CUBIC INCHES for a 4 stroke engine.
And More Stuff:
In case you ever wondered about the power reqd. to pull you car at a specified speed; here's some unsophisticated formulae: Pressure in pounds per Sq FT of Air resistance=.003 X MPH squared. Horsepower reqd. =.000008 X MPH cubed per Sq. Ft. for air resistance. Horsepower reqd. for rolling resistance = .0001 X Wt. of car X MPH.
Using the formula to find the HP for a car to run 200 MPH at Daytona; assume the car weighs 3500# and is 30 sq.ft with a drag coeffient of .25.
200 cubed = 8000000 X 7.5 sq.ft.= 60000000 X .000008 = 480 HP for air resistance plus 3500 X 200 = 700000 X .0001 = 70 HP for rolling resistance = a total of 550 HP.
I think we have heard the nascar teams talking about 550 or so HP with their restrictor plate motors.
More Stuff Coming:
Some years ago I was sorta skeptical when reading about 6-7,000 HP + in the Top Fuel dragster engines in cars that can cover the standing 1/4 mile in 4.44 seconds at 333 MPH. Some time ago I came up with a rough formula to try and confirm this HP rating.
To determine the force reqd. to cause a certain acceleration I used the age old formula F = M X A where F = force in lbs, M = mass & A = acceleration in ft per sec per sec.
Given that the car weighs about 2250 lbs for a mass of 70.
The tire is 36" tall for a radius of 1.5 ft.
Assume a coefficient of traction of at least 1.
The rear end gear ratio is 3.2 to 1 and there is no transmission used.
A slipper type clutch is used with variable lock up rpm/time.
Timing devices indicate that after 1 sec. the car is at 125 mph or 182 ft per sec.
The car has accelerated at a rate of 182 ft per sec per sec and may exceed this.
We will assume 6,000 rpm at the 1 sec time, but the clutch is not yet locked up.
Max. engine size is 500 CI.
From F = M X A we get 12,740 lbs force reqd. which we must multiply by tire radius of 1.5 ft = 19,110 which we divide by the gear ratio of 3.2 to obtain 5,972 ft lbs torque on the drive shaft.
At 6,000 rpm and 5,972 ft lbs the HP is 6,822 and let's not forget while this is going on at least 10 % of the total engine BHP is being lost as heat from the slipper clutch which can reach temps. of 1,000 degrees F +.
It is no longer a problem for me to believe the 7,000 HP they're talking about and if you see them live you KNOW it's true.
As a curiosity we find from the BMEP formula: (BMEP = Torque X 150/CI) that the effective average pressure on the power stroke is 1,792 psi. I don't think the Max. combustion pressure is as proportionally high as a high performance gasoline engine or we could expect around 5,000 psi and it couldn't stand that; so they run the engines extremely rich to hold the Max. pressure down. Actually the combustion may be closer to a "constant pressure" as in a theoretical diesel that an actual diesel.
This indicates a tremendous pressure in the cylinders and a very good reason for the Hemi engine to be best suited for this type of work. The combustion chamber is a slice from a sphere. Years ago while working in pressure vessel design I learned that a spherical shape is by far the most efficient design for pressure retaining parts.
One more pile:
It has became obvious that DE,Jr has a plate motor advantage; it has to be something unconventional and I'm wondering if they have maybe discovered that contrary to popular belief "shorter" con rods can make more power. Assuming the conventional engine has a stroke of 3.5" and a rod length of 6.125" so that the length to stroke ratio is 1.75. Compare this to the BMW M3 that has a ratio of 1.48 and that engine makes 1.7 hp/cu in. And likewise these little newer 4 banger auto engines have ratios at about 1.5.
Now consider the worlds largest, most powerful and most efficient engine has a ratio of about 1.16. Look at the "to scale" cross section and do a little measuring. And my guess none of these engines were built like this just to save height.
http://gcaptain.com/emma-maersk-engine/
10 Cyl, inline diesel, Sulzer Marine, 38" bore X 98" stroke, 100,000 HP @ 101 RPM
I have a ME handbook that has about 6-7 pages of charts etc. on crank operated reciprocating elements and with different rod to stroke ratios. The charts show piston position and velocity relationship to the crankpin at 5 degree intervals, angles for different fractions of the stroke, inertia factors and piston accelerations at 5 degree intervals. There is also a section about offset cranks and the really strange part is with an offset crank or wrist pin the stroke will be longer than normal.
Main thing to remember is the ratio of piston velocity to crankpin velocity is referred to as the tangential factor or crank turning ability. It is very easy to look at these charts and see that a shorter rod provides a higher T factor AND it come sooner in terms of crank degrees AND time. Time is very important with respect to blowby leakage and BTU leakage to the water jackets and nearly all of this occurs near TDC. To me this is total evidence that theorietically the short rod should makes more power. Most all say the extra angularity and friction loss is more than any gain of the short rod; I always felt the drag of the rings was the bigger part and angularity would not matter that much. On TV years ago I saw a dragster make several runs blowing oil smoke like fogging mosquitoes and he was the fastest in his class; I'd bet he had intentionally left the oil rings off to save friction.
Another strange thing is that at 90 deg. ATDC the T factors are all 1 regardless of the rod length.
Now to the intake stroke and of course the piston behaves the same way. I look at it this way; if it were possible the best thing that could happen is on the intake the piston moves INSTANTLY from TDC to BDC and simply resides there until time to start on the comp. stroke, but of course that's impossible so I'll go for the next best and that is get it away from TDC as quickly and with as much velocity as possible. You know about RAM effect and by the way have you ever wondered why Dodge chose to refer to their first V8s as RAMS; no it wasn't because they looked liked goats.
Ratio of piston velocity to crankpin velocity with rod/stroke ratios of 1.5 and 2 and at 15 degree intervals from TDC to BDC as follows: 15 ATDC .3425 & .3215, 30 ATDC .6464 & .6091, 45 ATDC .8786 & .8341, 60 ATDC 1.0168 & .9769, 75 ATDC 1.0539 & 1.0303, 90 ATDC 1.000 & 1.000, 105 ATDC .8779 & .9015, 120 ATDC .7153 & .7551, 135 ATDC .5356 & .5801, 150 ATDC .3536 & .3909, 165 ATDC .1752 & .1962.
The greatest differences occur at 60 ATDC in favor of the short rod and at 135 ATDC in favor of the long rod. At 135 ATDC the pistons in both have slowed to only about half of their peak velocity and due to the slow down, my thought would be that ram effect should already be occurring at this point or at least not being created.
The above values also determine the crank turning ability and the total pressure on the piston times the above values determines the tangential pressure on the crank at that point.
One of the funniest things that you could read about is the same folks that say long makes more power will also say that offsetting the wrist pin in the direction opposite crank rotation will give more power and an OS pin and a shorter rod do about the same thing; in that they give the piston more velocity earlier.
If one sets out to outengineer the M3 he's got one heck of a job on his hands and no way would they make the con rod 1/2" or so shorter just to save engine height.
And you know, I think this short rod deal would be more at home on a plate motor than anywhere else, relativly low rpm and the restriction should like the extra piston velocity.
'51 Hemi & 55 SBC, a class of their own