Outboard Torque

[CDave]

OK, we all know the HP rateings on engines but how much torque do the V6's make?

[MTCM]

For known horsepower (HP) at specified RPM:


HP X 5252 divided by RPM = Torque (ft. lbs.)


300 hp @ 6000 rpm

300 X 5252 = 1575600 divided by 6000 = 262.6 ft. lbs.


HP = Torque X RPM divided by 5252

Clear as mud huh? :D

[CDave]

Is that for 2strokes, 4strokes and rotories?

That really seems like a vague formula to me. Seeing how rod length, stroke, bore size among so many other things determine an engines torque. But torque seem to be a BIG part of an outboards overall perfromance, i.e. a Merc 2.4 vs. Merc 2.5.

I'm trying to learn about these outboards and all there ins and outs. I know 4stroke and 2stroke industrial engines but I'm new to 2stroke outboards.

[Mark75H]

That formula is for all motors, even steam engines and electric motors. The type of engine has nothing to do with the final math involved here.


rod length, stroke, bore size among so many other things

These factors may influence a particular motor's torque characteristics, but they do not get between horsepower and torque. If you have 100 ft lbs of torque at 5252 rpm you have 100 hp, if you have 300 ft lbs of torque at 5252 rpm you have 300 hp.

MTCM correctly used the formula to tell you the torque of a 300 hp @ 6,000 rpm motor.

When a motor is run on a dyno, the dyno does not directly measure horsepower- - -what it measures is torque. After that a person (or computer) has to calculate what the horsepower will be from the torque reading and the rpm using the lower formula MTCM posted:
Torque X RPM ÷ 5252 = HP.

[MTCM]

(Torque x Engine speed) / 5,252 = Horsepower

The engine that makes 300 pound-feet of torque at 4,000 RPM produces [(300 x 4,000) / 5,252] 228 horsepower at 4,000 RPM. But where does the number 5,252 come from?

To get from pound-feet of torque to horsepower, you need to go through a few conversions. The number 5,252 is the result of lumping several different conversion factors together into one number.

First, 1 horsepower is defined as 550 foot-pounds per second. The units of torque are pound-feet. So to get from torque to horsepower, you need the "per second" term. You get that by multiplying the torque by the engine speed.

But engine speed is normally referred to in revolutions per minute (RPM). Since we want a "per second," we need to convert RPMs to "something per second." The seconds are easy -- we just divide by 60 to get from minutes to seconds. Now what we need is a dimensionless unit for revolutions: a radian. A radian is actually a ratio of the length of an arc divided by the length of a radius, so the units of length cancel out and you're left with a dimensionless measure.

You can think of a revolution as a measurement of an angle. One revolution is 360 degrees of a circle. Since the circumference of a circle is (2 x pi x radius), there are 2-pi radians in a revolution. To convert revolutions per minute to radians per second, you multiply RPM by (2-pi/60), which equals 0.10472 radians per second. This gives us the "per second" we need to calculate horsepower.

Let's put this all together. We need to get to horsepower, which is 550 foot-pounds per second, using torque (pound-feet) and engine speed (RPM). If we divide the 550 foot-pounds by the 0.10472 radians per second (engine speed), we get 550/0.10472, which equals 5,252.

So if you multiply torque (in pound-feet) by engine speed (in RPM) and divide the product by 5,252, RPM is converted to "radians per second" and you can get from torque to horsepower -- from "pound-feet" to "foot-pounds per second.

[CDave]

Thanks y'all, I wasn't trying to say anyone was wrong.

[MTCM]

Addressing the 2.4 vs 2.5 thing. Because of the displacement difference the 2.5 will produce more horsepower at a given rpm than the 2.4

A 2.5 200 and a 2.4 200 both put out 200hp right? Most likely the 2.5 will produce it's rated power output at a lower RPM than the 2.4

[ShorePounder]

...and therefore produce more torque at the lower RPM correct?

I'm picturing this in my mind to help relate . A 3406C Caterpiller diesel is rated at 425 hp, but redlines around 2300 rpm. do the math and that's a truckload of torque. An LS6 454 was conservatively rated at 425hp (we'll use that number for this comparison) but at 2300 rpm's it's not wound out nor is it at it's peak so logically and mathmatically not as much torque.

I think I got now. :D

[Techno]

The other thing that hardly anyone considers is max power or torque aren't all that important.
Any engine can be built pretty much in 3 compromise levels.
Low end, mid range, high end.
Torquey, drivable, max power.
Notice I put drivable in the middle because I experienced a stump pullin low end torque cam in my 4:11 Vette-427. It wasn't a good combination. Hey it's a street car so why not develop the power down where it's driven?
How much power is developed at a certain rpm and whether it is flat or peaky is really what is important for any given use.

For the 2.4- 2.5 example. The 2.5 is detuned to produce more power lower or to maintain a cetain power level longer. It could produce more hp than 200 but then it would be borrowing from somewhere to do this.
The 2.4 could have more power in the low end or a flatter curve but then it wouldn't develop 200 hp.
If you put a 450 hp 2.5 260 on a given boat and slap on a 32" wheel it's top speed could be just under planing speed. It wouldn't develop the power needed to overcome this large power requirement at a low engine speed. So in effect this 450 hp engine would only develop say 60 hp. When it produced more it would do what my Vette did, the prop would cavitate and the power would be wasted, does nothing.
In my case it did accelerate the car. It was also a cop caller and spun the tires whenever you turned from a stop. It was too much power with no load to consume it properly. Just a real life example when you go to the other extreme from high rpm horse power to low rpm horsepower. Too much of an extreme is only useful in extreme uses.

BTW electric motors and steam engines both can produce maximum torque at zero rpm. They don't need gear boxes just an ample supply of external energy. This would be zero hp until the locomotive began to move, or the wheels spin.

[Russ]

Need to correct your figures from dyno pulls back to sea level standard (reffered Hp or Qe), to keep relevance between runs, engines, or different test condition ambients.

[CDave]

OK, I was at the Wakasha factory a few years ago and saw a 7042GSI being dynoed.

Waukasha rates these engines at 1200hp@1200RPM @5lbs of boost.

The dyno screen showed 1200rpm's, 5lbs boost, 1225HP and 4752ft lb of torque. What gives? Is it because Waukasha, Wisconson is above sea level?

Shorepounder do the math on a Cat 3616G, 4235hp@900rpm. :eek:

[Russ]

after the fact. The mfg states a power rating based on something, and would assume it's a sea level std. rating with the appropriate SAE correction factors applied to their raw data. Don't know what type of dyno system was being used, some are very sophisticated and have test cell ambients plumbed into the software and perform the correction factor calculation and display corrected numbers. Other may have to be measured and corrected out manually. 1225 Hp is within around 2% of the claimed rating and probably within the mfg's production spec and accounting for any inaccuracys with the dyno. Who's Shorepounder? Oops, never mind...saw him above.

[Mark75H]

Dave, you are saying that the 4752 ft lb @ 1200 rpm should equal 1085 hp, correct? Well you are right, that motor was only actually making 1085 hp there that day, even though the screen was showing 1225 hp. The 1225 hp was "corrected hp"

"Corrected hp" corrects for differences in intake air temperature, humidity and barometric pressure. If the motor at Wakasha was running at all the calculated "ideal" parameters it would have been pulling 5361 ft lbs of torque to pull 1225 hp @1200.

The variance of the conditions (from std ideal conditions) in the dyno cell allowed a 12% increase in calculated torque to be used to calculate the hp.

Some of it would be because of the lower barometric pressure normal at Wakasha because of the altitude, but much of it would be temperature and humidity correction factors.

[CDave]

Thanks agian. :)

[NPK2003]

Dang, its hard to be from Wisconsin with so many Indian derived city names, but it's Waukesha, Wisconsin. Same city my business is in. We sell and service Waukesha Engine machine tools. So, Dave, if you're up this way again, I'm 5 minutes away from their plant. Let's have lunch or something?
;)
Nick

[CDave]

Originally posted by NPK2003
Dang, its hard to be from Wisconsin with so many Indian derived city names, but it's Waukesha, Wisconsin. Same city my business is in. We sell and service Waukesha Engine machine tools. So, Dave, if you're up this way again, I'm 5 minutes away from their plant. Let's have lunch or something?
;)
Nick

Sorry there, grammer has never been a strong point of mine. Try living in Calcasieu Parish with my bad spelling. :D

I really don't like the Waukesha 7042GSI's, head crackers as I call them, so maybe that is why I never spell it correctly. :p

If it ain't a Cat it's a Dog. :D J/K