Nitrous & torque

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EF9_boy

New Member
how does nitrous ratio hp/lbs-ft ? i know a 50hp shot is an extra 50hp but what kinda torque would that result in?
 
how does nitrous ratio hp/lbs-ft ? i know a 50hp shot is an extra 50hp but what kinda torque would that result in?
I would think that would depend on several factors including the rotational mass of the engine/flywheel.
 
ok a stock b series flywheel weighs something like 18lbs lets say its on a b20z which produces 146hp & 133lbs/ft at the flywheel so with a 20% loss through the flywheel it should make roughly 117whp so if you add in a 50hp shot it would make est. 196fly or 157whp
so with all that factored in (if i am on track) how would you calculate torque?
(other than a dyno lol)
 
Peak torque? Nitrous, at least for a steady shot(not pulsed like injectors) will result in about the same HP gain throughout the power band, so you can work backwards....which means you need the RPM.

HP=RPM*Tq/5252

Assuming peak torque is at 5000rpm, you'd have (126.6hp*20% loss= 101.2whp+50whp shot=151.2hwp=158ft-lb/tq, or an increase of about 58lb at the wheels at 5000rpm. This is of course, all theoretical. Actual numbers will vary; though you'll usually see larger gains in the lower RPM and it tapers ass RPM rises(hence why overall HP gain stays about the same).
 
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101.2whp+50whp shot of nitrous=151.2whp @5000rpm(the RPM we're assuming).

151.2whp@5000rpm= 151.2*5252/5000= 158.8lb of torque, at the wheels.
 
Also, keep in mind that the 1 shot = 1 hp rule of thumb is for wet shots and direct port set ups. It's not for dry shots, IIRC.
 
you have to put it on a dyno, every engine is unique. 50 shot might be 40hp in one and might be 50-60hp in another.

never take the shot number as hp numbers, take is as a way to measure how much nitrous is going in the engine.
 
That wasn't his question; it was more of how would you find out the torque increase. In this case it's an estimate. Most nitrous sizing is based off of HP-estimates.
 
Yeah I see that, but there is no real way to estimate horsepower/torque numbers accurately. You might as well just throw a number out there. Only real way to find out is Dyno.
 
for rue numbers magana559 is right. a dyno is needed or you can just guess. its like the engine in my avitar. thats a 302-- 5.0 for my 82 mustang GT. its an 87 year engine set up with a 4-barrel carb, a 200 shot of nitrous on a plate system with full throttle switch. at best geuss with the stock hp and tq figures plus the extras, its puttin about 550 hp. however, the car is running just 245 tires and a 3:27 gear ratio inthe rear end. that power dont even hook up good. but i can do john force style burnouts great!! it might push a dyno to 375 hp but its a guess
 
Uhh...

Ask Calesta?

Honestly, I've never played with it.

you have to put it on a dyno, every engine is unique. 50 shot might be 40hp in one and might be 50-60hp in another.

never take the shot number as hp numbers, take is as a way to measure how much nitrous is going in the engine.

Yup. Nitrous "shot" is just an estimate based on horsepower increase- but all it's really doing is chilling the air to increase density and adding oxygen when the N20 breaks apart. I guess you could use some of the previously mentioned math to work it out, but the best way to check it is to get the car on a dyno and see what happens.
 
this is an old thread but i figured i'd clear things up.

nitrous ALWAYS produces more torque than horsepower this is a fact no matter how many cylinders you run.

a shot usually equates to WHP, not crank.

i read about a honda s2000 with bolt ons running a 75 wet shot.

he made like 70 RWHP and 90ftlb's RWTQ.

with a V8 its a whole different ball game.

this is a generic example but a 100 shot on an 8 cylinder would give hime around 100 RWHP and anywhere from 170- 200 ft lb's of RWTQ.

nitrous is the shit.
 
i have always been told that as well... that nitrous makes more torque than hp. im not real big on mathematical calculations so ill leave it at that :p
 
nitrous ALWAYS produces more torque than horsepower this is a fact no matter how many cylinders you run.

That is incorrect(unless maybe you're talking about peak figures?); again it's basic math. Hint: you'll see the most drastic difference in high-rpm engines.
 
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