Displacement -vs- HP

The characteristic that is generally directly proportional to displacement is torque. In the example above the 502 would make approximately 25% more torque than the 396 and at a lower RPM. It is also worth noting that while the 502 with the same heads etc. will not make much more power than the 396 its peak horsepower will occur at a lower RPM.

The bottom line is that, all other factors being equal, torque is directly proportional to displacement, but horsepower is not necessarily. In many tractor applications where the engine is operating at such a low RPM that it is probably capable of nearly 100% volumetric efficiency (ability to flow air at a high enough rate to fill the cylinders 100%) throughout its entire operating range you may see directly proportional HP gains with displacement increases up to a point. There will come a point as the engine gets bigger and/or you run it at higher RPMs, however, where your horsepower will be limited by your cylinder heads, intake and exhaust manifolds, and camshaft.

I really enjoyed and appreciated your post. I just wanted to add that the cylinder heads on these old tractors are so poor that the typical "factory" maximum volumetric efficiencies are in the mid to upper 70's. When the induction system is improved (through head, cam, and manifold work), it still only reaches the 85 to 88% range. However, even this modest improvement in volumetric efficiecy, which is often accompanied by very substantial increases in displacement, add up to quite drastic improvements in power.

Regards,

Goldsburg

Keep in mind that factory carburetors weren't sized for maximum power! They were sized for driveability, economy, responsiveness, and reliability. All of these factors will lead you to a carb that is smaller than ideal for maximum power. With that in mind, imagine what the ramifications are when the factory cubes are doubled!

The premise of this mod is that the engine vacuum signal is SO strong on modified antique tractor engines, even at wide open throttle, that the pressure drop required for proper fuel flow occurs through the entire carb. This eliminates the need for a "special component" called a venturi, because the entire carb has now become the "venturi"! Compare the diameter of your throttle plate to the diameter of the carb opening. This might remind you of the way that the venturi has one diameter for an opening and a smaller diameter in the middle...?

The main fuel circuit components will need correspodingly enlarged because the signal is not as strong as if the original venturi were in place, but it is certainly strong enough to meter fuel, if the correct components are modified.

In addition, dyno numbers do not lie! I have personally witnessed (on my own tractor) a 15% gain in power from this modification. This is quite substantial, especially considering that the carb is what was "legal" for that class. When there is no carb limit in a class (stock appearance doesn't matter), then this particular mod is not necessarily right for you or anyone else in that circumstance.

The fundamentals of physics are what makes carburetors work (thank you Bernoulli!). This mod was developed using those principals and is completely sound, viable, and reliable. Just because that you do not understand the way or the reason that some things are done doesn't mean that they are wrong....

Regards,

Goldsburg

I appreciate your commentary, however the dyno work that I have done is on stock +XX% RPM tractors.

You have your theory about how carbs (and engines)should operate and I have mine: I personally do not want to have lug my engines down to get them to make power, for two reasons: Loss of ground speed and loss of airflow (power). If you have a governor on an engine, that governor will keep that throttle plate nearly closed until the load hits. The governor will then adjust the throttle plate position correspondingly to the engine load. Once the throttle plate reaches the "wide open" position, you have maxed out airflow and thus power. Any RPM reduction beyond that means that you are reducing airflow (and power). Your torque may increase as you drop in RPM's, but it will not increase as fast as the RPM's are dropping. Also, as torque rises you are now much more likely to spin out or loose traction in a loose spot.

The fact that you have reached and are now below max power levels means that you are no longer putting maximum power to the ground. If you are dragging your engine down well past the point that the throttle plate is at the "wide open" position (as determined by the governor) then I would suggest that you are in too high of a gear!

How many Open class pulls are won by the guy that has lugged his tractor waaaaay down...?

Regards,

Goldsburg
goldsburg@att.net

I will try to slow down a little. Stock RPM +XX% is read as "Stock RPM+ (whatever)% (RPM) over". This refers to stock RPM +20% over, stock RPM +10% over, or stock RPM + 30% over, etc. Have you ever heard of the NATPA or USAP? That is one of the rules that both of them (and most other antique tractor pulling organizations) have. Get the picture?

No, I don't win every pull. But I do observe the other pullers (going down the track) when I am not pulling. I do not stand around an BS with other pullers and then have to run to my tractor when it is my turn to pull. I am sorry that I am what you might consider a fierce competitor by not BS'ing, but I like to win and enjoy doing so.

Why are you so offended that I do not agree with you? As I said in my prior post, you have your theories and I have mine, end of story.

As far as you being old, I do not know about that at all. What I CAN tell from your posts is that you are too stubborn to even consider other the side(s) to an issue...

Regards,

Goldsburg
goldsburg@att.net

I rather doubt that in all engines the torque and horsepower cross at 5252 rpm. Are you telling me that a 900 cubic inch "G" at say, 180 HP, and a Thundercat at 180 HP, both will have matching torque and horsepower at 5252 rpm? I think you would agree the "G" would be history at that speed and the big Cat would have fallen on it's face at such a slow speed.

The high-powered sled engine and the monster 2-cylinder engine both would do exactly the same amount of work at equal horsepowers. When the R's start dropping, the "G" comes on strong in the torque department, which, by the formula, keeps the horsepower from falling off rapidly. The Cat on the other hand, with it's "peaky" torque curve, will lose horsepower rapidly once the R's go away. Rest assured, at their power peaks, both engines are work output equals!

The original post was something about "no replacement for displacement". I agree that increasing displacement will most certainly increase power output. By itself, this simply increases the amount of air/fuel that can flow into the power/cylinder, be compressed and ignited. I believe much more power is gained by increasing compression ratio and rotating speed than by merely overboring the cylinders.

"Different strokes for different folks"

This formula is used to determine horsepower at a given speed and a given torque. Maximum horsepower is not synonymous with maximum torque.

And to reply to another post, torque as a measure or rotary force in the modern day formula and the foot-pounds term used in the original horsepower formula as a measure of linear force are both very accurate measures used in determining power output of any power unit, be it iron or be it horseflesh (or spaceshuttles). It's just that all power units are not capable of being measured in both manners.

I'm assuming that we aren't talking about a Div 5 tractor or anything near it, since that didn't seem to be the case with the original post. I presume he is pulling Div 1 or Div 2.

Belt HP. and Drawbar HP.

Pulling a sled is the classic example of HP measurement. If you had a spring scale instead of a chain hooked between the sled and the drawbar, then figured the weight pulled a certain distance and measured the time it took, you have HP!
The confusion, I think, comes from people incorrectly using the term "Torque" to describe the power that is coming from an engine.

I didn't mean to start an argument here. I was just hoping to share some information.

Time is an essential number in any horsepower formula. If time isn't applicable, then how do you compute horsepower?

I agree with your statements, which are based in Physics and real life. However, I have found that trying to have a meaningful, thought provoking, and scientific conversation with the likes of people (like Earl "SEAMO") would be about as productive as trying to teach a dog to fart.

Thanks for trying though! I am sure that there are other people reading this will appreciate (and understand) your comments...

Regards,

Goldsburg
goldsburg@att.net

Unless you are given extra points for moving a sled faster, and that would be hard to determine, since the sled weight varies as distance increases, you are better off to know you have enough torque to do the job.

If it had lots of $$$, and could engineer my own engine, I'd make horsepower one big consideration. But in real life division I and II pulling, you must stick pretty much with what the manufacturer puts in that tractor. As one of the original posters said, once you've slowed the engine to where the volumetric efficiency is high, you'll be getting all you can from that engine. This will not be at the point where you are making maximum horsepower.