Sweetshoes90210

Hey. Dose anyone know how to figure out the footpounds from the CEN .16 nt engine stock and then figure out what the footpunds are with all the gears

braden.r productions

Footpounds of torque????

cummins driver

My Feedback: (31)

Its not much. Im not sayingthe motor, is weak, im just saying that i dont think it would be much on any nitro motor, unless its pretty mean. Youd probably be better off measuring in inch pounds.

Eric

mecky33

Cant find a definite answer online, I will call Cen tomorrow.

slodsm

My Feedback: (1)

Standard units of measure for tq are ft lbs and in lbs. Hobby engines are rated at oz in of tq so it would be very small. I am a mechanic by trade and don't even know how to convert that measurement because we don't use anything that small.

Sweetshoes90210

well mecky any relpy from CEN?

mecky33

They don't really have any specs on the NT-16. Mostly because there are so many variables that affect torque output. The closest I could find using a torque calculator designed for real cars said at 25000 RPM it was putting out about .25 foot pounds.

vizaracer

what engine is that??? do you know how much my sirio on my signature will be putting out??

cummins driver

My Feedback: (31)

I guess that would be about 3 inch pounds then.

Eric

mecky33

Like I said Visa, its all relative to what pipe, fuel, filter, etc. is on your truck. Just find out what your Peak HP rating on your engine is and what RPM its at, the do a search for a torque calculator. There are a bunch of them on Mustang, Corvette, and other performance car dedicated sites. Follow the directions on how to use the calc and go from there. That will give you a base number to go by. And the NT-16 is a Cen factory engine.

joules

hp = ((torque*rpm)/5252)

conversely

(hp*5252)/rpm) = torque

rough calculations of a .21 motor are
(2.2 hp*5252)/30000 rpm = .385 ft/lb of torque

.385 ft/lb * (12 in/lb / 1 ft/lb) * (16 in/oz / 1 in/lb) = 73.9 in/oz roughly at peak hp
J~

petrolpixie

hmmm you lost me there lol

nitromacguyver

i see wht you did. theres only one problem, the engine in question is a cen mt2 with a .16 size engine that is said to put out 1.2hp.

so if i take your formula and crunch some numbers i will get...

((hp*5252)/rpm) = torque

((1.2hp * 5252) / 30000rpm) = .21008 foot/pounds of torque

and then convert that to inch/ounces of torque...

.21008 *12 inches per foot = 2.52096 inch pounds * 16 ounces per foot = 40.33536 inch ounces of torque

and you get;
.21 foot pounds of torque,
3.36 inch pounds of torque,
or 40.3 inch ounces of torque.

there you have it nice and simple.

mecky33

nitro, the RPM should be 25,000. The formula requires the RPM at which the engine creates peak HP. In the case of the Cen NT-16, it is 25,000, with a max RPM of 33,000.

Sweetshoes90210

Cool, thanks...

vizaracer

i figured out that i am running about .43 foot pounds of torque. i am running 5.05 inch pounds of torque and i am running 80.68 inch ounces of torque. am i calcualting this correctly?? i have a 40,000 rpm engine. with about 3.2 HP. i put the turbo plug in it also. and i am running the motor saver air filter and i am running the HPI ribbed tuned pipe.

joules

Peak torque usually occurs at a different RPM then peak HP. Also, peak hp does not occur at max rpm.

On an O.S VZ-b(p) it turns 3,000-40,000 rpm while peak hp (2.45 hp) only occurs at 32,000 RPM. Up at 40,000 RPM you may only be running 1.8 hp. While max torque may be produced at only 27,000 rpm. This is just some guesses as they only state max hp at an rpm and usable rpm range on their site.

http://www.osengines.com/faq/faq-q686.html
J~