couple b questions

What is the entire engine #. Could be R1##### which would have been a power unit motor.

According to one book the power unit motors had R or CR or the standard CE prefix, all 125 CI motors.

Depending what the power unit came off of it may have a shorter govenor spring to give you more RPM. I have 2 tractors with power unit motors on them. 1 is the normal CE set up but the other one , from a combine, has a shorter spring on it and really moves on the road. Bob

The springs are on the govenor and control the RPM. Bob

So can u just put a better spring on a c engine to get more power out of it?

So torque would be your main focus if your trying to get more power? Because torque is what actually moves the tractor right?

5252 is a constant that never changes... kind of like PI = 3.14 when you figure the diameter of a circle.. And Dave is right, you want more torque, or the same torque at more RPM.. A dyno measures twist on a shaft or "torque" and then it multiplies by the rpm and constat to get the HP.. HP is just a calculated number.

Edited by steve(ill) - 18 Feb 2012 at 9:12am

5252 is attributed to the ability of an 18th century English mining pony to pull a rope.

One mechanical horsepower equates to 33,000 foot pounds of work per minute...

That is, the ability to raise one pound of weight, 33,000 feet, in one minute...

or lift a 33,000lb weight one foot, in one minute...

Or a 550lb weight one foot, in one second...

which certainly doesn't immediately identify where the magic number 5252 comes from, but...

Take a wrench with a handle one foot long, and pull on it with a force of 525 pounds... and you have the equivalent torque of 525 foot pounds.  If you can spin that wrench for about a sixth of a rotation, in one second, you'll have proven yourself worthy of producing one horsepower.

Why a sixth of a turn?

Well, you're turning a  wrench that is one foot long... that means the DIAMETER of an equivalent drum is 24" across, which has a CIRCUMFERENCE of about 75 inches.  A 75" drum will lift that weight SIX FEET in one rotation.

How's this relate to 33,000 foot pounds?   Take 5250 and multiply it by SIX.

A 2' diameter drum, spinning at 5250rpm, will pull 33,000 feet of rope in one minute...

(And for purists, the reason why the math doesn't come out exact, is simply because I truncated the decimal places.)

and with one foot-pound of torque applied to the drum, you'll have one horsepower.

Here's the detailed math:

Drum with 12" radius = 24" diameter, or (24Xpi) = 75.398223686155037723103441198708 circumference or
divided by 12 inches = 6.283185307179586476925286766559 feet.

6.283185307179586476925286766559 * 5250 =

32986.722862692829003857755524435

Plus or minus a little because the math was done on a binary computer Powered By A Battery, but running mathematical process on an INTEL processor, running Microsoft Calculator.

But I digress...

It all comes back to an English mining pony... and a guy named James.

AJ wrote: So torque would be your main focus if your trying to get more power? Because torque is what actually moves the tractor right?

"Power" is subjective, and without a very clear definition of what goal one is trying to reach, modifications for 'more power' won't necessarily result in the goal.

On industrial engines (stationary), the power ratings imply things WAY beyond pulling that rope.

See, James Watt was developing a steam engine, and it had a pretty big bore, and a pretty big stroke, and although it couldn't spin really fast, it could pull the rope across a large drum, and it could run a pump to keep the mine from flooding.

Imagine the amount of force that a 6" piston can exert when 100psi of steam is applied... pretty simple, really- calculate the area of the piston (6" diameter = 3" radius, and Area = Radius x Radius x pi), so we're talking about a little over 28 square inches of piston surface.  With 100 pound of pressure against 28 square inches, you've got 2800lbs of linear thrust.  If that piston is pushing on a crank that is 1' from center to throw, you'll be getting 2800 foot-pounds of torque. 

(Extra credit Question 1:  If a steam engine generates 2800 foot-pounds of torque, ABOUT how many rotations per minute does it take to replace ONE English mining pony??)

(Extra credit Question 2;  If that steam engine turns at 400rpm, how many English mining ponies can it replace?)

Now take a C engine that has been running with the same internal components for 25 years, add a heavier shorter governor spring, crank it up and expect way less horse power when the engine locks up because the rings slammed into the ridges at the top of the cylinders and broke up causing a malfunction. That may be like trying to enter a 18th century English mining pony in the Kentucky Derby and expecting it to be alive at the end of the race.

In a stationary and agricultural environment, there's much more than foot-pounds and RPM.  As Charlie notes, an English mining pony wouldn't last too long in the Ky Derby (but it'd be fun to put 'em in a pen together- that mining pony would bite the Derby horse like a piranha in a toilet bowl full of hamburger)...  nor would the Derby winner fare well pulling a mining rope.

But the point being... developing numeric horsepower under one set of parameters, doesn't mean it'll be worth a crap under others.

I can easily prove to you that A well-worn 60-year-old WC engine will outpull a freshly-built hard-block'd 2000hp supercharged nitromethane engine any day of the week...

By putting it out in a 40-acre field for three hours.

The hotrod motor will melt before making a turn at the first fencerow.

And finally, it doesn't matter how much horsepower you have under the hood, the limiting factors of TRACTIVE EFFORT, are the diameter of the driven wheels, the amount of drag imposed by all wheels, and the amount of weight applied to driven wheels.  Basically, you can simplify Agricultural Tractive Effort (aka 'drawbar horsepower') to being limited by weight, tire diameter, and drag... and once you've reached the effort threshold (where wheel 'slip' exceeds about 6%, that the amount of input horsepower required to pull harder, goes up by a very large ratio... hence, pulling efficiency under wheelspin becomes incredibly inefficient.

Power development, and converting it to tractable power, are two totally different things.  If you're looking to pull a plow that rolls your soil best at 3mph, and your tractor spins wheels, you'll need to either lose drag, or increase tractive effort by adding weight or increasing tire diameter.  Reducing drag means either changing tire size on non-driven wheels, or increasing tire diameter.  If you increase tire diameter, you'll have to reduce engine speed to stay at 3mph.  If you increase governed speed to increase engine horsepower, you'll have to change gear ratios to maintain your 3mph ground speed.

Managing and optimizing drawbar horsepower, as a result, has very little do do with engine horsepower... it's about balancing the draft load, against the tractor's weight and drag characteristics, against the soil's type and conditions, and finally... being able to do it for a continuous length of time under wide ambient and other variable conditions, including, but not limited to hot days vs. cold days, high and low humidity, dusty conditions, low quality fuel, fuel contamination, and all sorts of other little things.

But remember- it's a FARM TRACTOR.  It was built to be a farm working machine for everything from heavy tillage to operating elevators and augers, cutting and baling, and things that the original engineers could have never envisioned.  In order to 'tailor' it to suit a specific need, SOMETHING will have to be given up... and when doing such tailoring, it's good to have a VERY clear concept of what one is attempting to achieve, and realize that in ANY change, there will be a consequence.

In terms of drawbar effort, that's exactly right.  The engine governor limits only the upper RPM limit, nothing more.  The 'horsepower' figure of a higher governed RPM is useless when the engine's torque limit is found at a lower RPM.

The only way to take advantage of the higher calculated horsepower of a faster-spinning engine, is to use lower gear ratios, and if you can't find the appropriate ratios to allow the engine to stay at that speed, (or you can't keep the tire slip ratio low enough by virtue of weight/balance/tire diameter) there'll be absolutely no benefit.

The biggest failure many make, is to try and modify the engine to produce a 'peak' - a point in the powerband where tuning circumstances cause the engine to develop high amounts of torque at a given RPM range.  This is great for a sportscar or motorcycle, but when applied to a drawbar implement, becomes an incredible hinderance, because as soon as the torque required to continue the pull rises above engine's available torque, it pulls the engine off the 'peak', and you're done.  Pointless to have a 'peaky' engine in a farm field- better to have a 'flat' torque curve, or better yet, one that generates more torque as the RPMs fall, so that when a high-load portion of the field is in the plowshare, the engine's speed may come down a bit, but the available torque comes up.

Can't remember what engine it was, but I was working on a big natural-gas fired beastie many years ago, and it was horsepower-rated for it's max goverened RPM (wanna say 1200, as I think it was driving a 480v generator), but after looking at the engine's torque ratings, it had almost twice as much torque available at 100rpm BELOW it's low-idle point.


Edited by DaveKamp - 18 Feb 2012 at 3:05pm

Answers to Extra Credit questions:

1) about 12rpm
2) 33 ponies