Got the turd on the dyno today

[-HIC-]

Good catch, I did not notice it~

[RedHeartbeat]

Is that just a pheonomenon related to how HP is calculated from TQ values?

[-HIC-]

This part might not be of interest to most readers, but several people have asked:
"OK, if HP = RPM x TORQUE ÷ 5252, then where does the 5252 come from?"
Here is the answer.
By definition, POWER = FORCE x DISTANCE ÷ TIME
Using the example in Figure 2 above, where a constant tangential force of 100 pounds was applied to the rotating handle at a 12" radius, we know the force involved, so to calculate power, we need the distance the handle travels per unit time, expressed as:
Power = 100 pounds x distance per minute
OK, how far does the crank handle move in one minute? First, determine the distance it moves in one revolution:
DISTANCE per revolution = 2 x π x radius
DISTANCE per revolution. = 2 x 3.1416 x 1 ft = 6.283 ft.
Now we know how far the crank moves in one revolution. How far does the crank move in one minute?
DISTANCE per min. = 6.283 ft .per rev. x 60 rev. per min. = 377 feet per minute
Now we know enough to calculate the power, defined as:
POWER = FORCE x DISTANCE ÷ TIME
so
Power = 100 lb x 377 ft. per minute = 37,700 ft-lb per minute
Swell, but how about HORSEPOWER? Remember that one HORSEPOWER is defined as 33000 foot-pounds of work per minute. Therefore HP = POWER (ft-lb per min) ÷ 33,000.
We have already calculated that the power being applied to the crank-wheel above is 37,700 ft-lb per minute.
How many HP is that? (37,700 ÷ 33,000) = 1.14 HP.
Now we combine some stuff we already know to produce the magic 5252. We already know that:
TORQUE = FORCE x RADIUS.
If we divide both sides of that equation by RADIUS, we get:
FORCE = TORQUE ÷ RADIUS (a)
Now, if DISTANCE per revolution = RADIUS x 2 x π, then
DISTANCE per minute = RADIUS x 2 x π x RPM (b)
We already know
POWER = FORCE x DISTANCE per minute (c)
so if we plug the equivalent for FORCE from (a) and distance per minute from (b) into (c), we get:
POWER = (TORQUE ÷ RADIUS) x (RPM x RADIUS x 2 x π)
Dividing both sides by 33,000 to find HP,
HP = TORQUE ÷ RADIUS x RPM x RADIUS x 2 x π ÷ 33,000
By reducing, we get
HP = TORQUE x RPM x 6.28 ÷ 33,000
Since
6.2832 ÷ 33000 = 1 ÷ 5252
therefore
HP = TORQUE x RPM ÷ 5252

Note that at 5252 RPM, torque and HP are equal. At any RPM below 5252, the value of torque is greater than the value of HP; Above 5252 RPM, the value of torque is less than the value of HP.

Power and Torque: Understanding the Relationship Between the Two, by EPI Inc.

[RedHeartbeat]

Is that a yes?

[FLATOP08]

Thanks for the great info. Still wish I had more torque. Any suggestion on another cam.

[-HIC-]

Heh, yeah man!! sorry to cut and paste all that info, but just wanted to put it out there.

[-HIC-]

My Torque # is over 300 from the punch on the dyno, peaks around 340 at 4500 and stays above 300 until 5300. if that helps. If you want extra torque without changing cams throw a 100 shot on there.
Bryan

[Wtfan18]

so you think my POS will push roughly the same #'s as your truck or what?

[-HIC-]

If you have headers, full exhaust and a good tune it should!

[FLATOP08]

Do you have a dyno sheet I could see? I have to be careful with to much power cause I still have the stock tranny with just a shift kit. I don't know where you live but like Ramsey I plow with my truck and tow alot. But dam I still want to go fast. lol