We have had a number of threads about which engine to pick where the most common response was to look for the best power to weight ratio you can get. There have also been a good number of posts advocating getting to most torque you can get... There is a current thread running about horsepower but we have not had a good discussion on horsepower versus torque in a while. This is not a thread about Cat Versus Cummins or Ford versus Chevy! The intent of the thread is provide food for thought for those folks thinking about how much "power" they need.. Torque, Horsepower, and gear ratios all conspire to meet the challenge of weight balancing performance and fuel economy.

There are many on the forum that already know more about these issues than I ever will. This post is aimed at those that are at the beginning of the learning curve. The example engines I used were the ones I have some experience with there may well be better examples. Hopefully those in the know will help the rest of us with some practical insights, that will aid those embedded in the selection process.

The Theory
Torque tells you how much force the engine puts out.....how much weight it can move. Horsepower tells you how quickly it can move it

If you have a force sufficient to lift the one pound of weight one foot, then one foot pound of work will have been done. The time it takes to accomplish this is where HP comes in.. There are three variables to consider Force, work and time

If that event takes a minute to accomplish, then you will be doing work at the rate of one foot pound per minute. If it takes one second to accomplish the task, then work will be done at the rate of 60 foot pounds per minute, and so on.
A fellow named Watt and concluded a long time ago that the average horse of the time could lift a 550 pound weight one foot in one second, thereby performing work at the rate of 550 foot pounds per second, or 33,000 foot pounds per minute, therefore 33,000 foot pounds per minute of work was equivalent to the power of one horse, or, one horsepower.

For purposes of this discussion, we need to measure units of force from rotating objects such as crankshafts, so we'll use terms which define a "twisting" force, such as foot pounds of torque. A foot pound of torque is the twisting force necessary to support a one pound weight on a weightless horizontal bar, one foot from the fulcrum.

If we rotate that weight for one full revolution against a one pound resistance, we have moved it a total of 6.2832 feet (Pi * a two foot circle), and, incidentally, we have done 6.2832 foot pounds of work.

Since 33,000 foot pounds of work per minute is equivalent to one horsepower. If we divide the 6.2832 foot pounds of work we've done per revolution of that weight into 33,000 foot pounds, we come up with the fact that one foot pound of torque at 5252 rpm is equal to 33,000 foot pounds per minute of work, and is the equivalent of one horsepower. If we only move that weight at the rate of 2626 rpm, it's the equivalent of 1/2 horsepower (16,500 foot pounds per minute), and so on. Therefore, the following formula applies for calculating horsepower from a torque measurement:

Horsepower = Torque * RPM/5252

If you want to play around with the numbers... just take your torque and multiple it times the rated rpm and divide the result by 5252 that will tell you what the horsepower your producing at that given point...

Other Factors
Torque is how much force you can apply and, consequently, how steep a hill you can climb. Horsepower is how many times a minute you can apply the force . The more times a minute you apply the force , the faster you'll go. But, just like pedaling a bicycle, the faster you go, the less full force you'll be able to apply with each stroke.

The faster a piston engine turns, the lower its efficiency. First is the inertia of the pistons going up and down in the boresâ€”this change of momentum uses up energy on each stroke. Secondly the friction of all the moving parts. And third the work to get air into the cylinder and then push the exhaust out.

The gearing ..Transmission and rear end ratios are designed to render the peak performance at some prescribed point on the torque curve.... Some manufacturers will jiggle the ratios to achieve better fuel economy at the expense of acceleration ability...
Common gearing for Diesels is a 4:63 rear and for gas its 5:38....I have seen some extreme variations with a low a ration on the big diesels of 4:11 and up to 5:88 on some of the heavier gassers.

The number of gears available in the transmission is also a factor... The more gears the better.

What does it mean?

As an example, the ever popular V10 the peak Torque is 425 at 3250 RPM'S and Peak hp is 310 at 4250 RPM's as compared to the popular Cat 330 which is 860 ft lbs of torque at 1440 RPM's...and 330 HP at 2400 RPM'S.
Using the constant of 5252 at 1440 RPM'S the Cat 330 is producing 230 HP at peak torque versus the V10 at peak torque which is producing 248 HP

If we change the diesel to a Cummings 350 then the performance pictures starts to change. The 350 will produce a 1050 ft lbs of torque at 1400 RPM'S and 350 hp at 2000 RPM'S.... Comparing it to a V10, the 350 is 280 hp versus the 248 hp for the V10. If we change the diesel to the ISL 400 rated at 400 HP at 2000 RPM'S then you are talking about 285 HP at peak torque at 1250 RPM'S in the case of the ISL 400 you are talking about 2.8 times the torque than the V10 and 15% more hp but the hp is a wash due to the rear end ratios which are typically 16 % higher on a gas coach... so you are down to the issue of weight versus available torque... Using a weight of 20500 for the gas coach with the V10 and 33500 for the diesel with the ISL 400 You are talking about moving 60 percent more weight with 280 percent more torque...

My coaches GVWR is 33500 with 1200 ft lbs of torque available to push it, my buddies coach GVWR is 20500 with 425 ft lbs available to pull it.. 33500/1200 translates 28 pounds of weight for every ft lb of torque versus my buddy having to move 48 pounds for every ft lb of torque... In my example of my buddies coach versus mine if he had the 860 ft pounds of torque from a cat 330 with roughly the same Hp as the V10 his power to weight ration would be 24... with the same gearing than he currently has I think he would have himself a hot rod.....

If we use the same weight coach say 23500 GVWR and compare the 330 cat to the V10 since they have similar HP rating at their respective peak torques. The Cat at 860 ft lbs of torque has a torque to weight ratio of 27 meaning for every ft lbs of torque the cat has to move 27 pounds of weight versus the V10 having to move 55 lbs...While some of the offset in torque can be dealt with gearing... The bottom line is that its the torque that is the deciding factor on terms of how much "power is available to move the coach......

Doing test drives, I drove four coaches that were comparable in weight, two were equipped with a Cat 330 HP with 860 ft pounds of torque the other two had a Cummins 330 HP rated at 950 ft pounds both had the same 6 speed Allison. The coaches with the Cat came equipped with a 4:63 rear and the Coaches with the Cummins came with a 4:30 rear. There was virtually no material difference in the timed acceleration runs that I did.. 0 to 45 and 45 to 60 MPH. I do not know if the lower ratio netted increased fuel mileage. I also don't "know" if the engine with the higher torque would have shown better results in my timed run had the rear end ratios been the same. While the inference of the increased torque would lean one in that direction, it would be speculation to assume is that there would have been a difference.

Conclusion

The heavier the coach the more horsepower and torque is needed to deliver acceptable performance. Up to a point available horsepower and torque can be leveraged by increasing the rear gear ratio. The down side is fuel economy. So when you are focusing on the power part of the decision equation you will need to concern yourself with the weight of the coach plus if you are going to tow, the weight of the dinghy then contrast that with the rated horsepower and torque and rear gear ratio.

The best way to determine if the power is adequate for your needs?? Test drive the coach. Keep in mind that your test drive will the lightest the coach will be and no dinghy in tow so in terms of performance it will be as good as it gets. Each of us has our own threshold in terms of how important performance is and when enough is enough. I found the timing of acceleration from 0 to 45 MPH and the then 45 to 60 to be a useful and recordable indication of performance.

Johnny T,
Thanks!! That is a very well thought out way to present a very complex proposition to us that need it in an easy to understand manner. I only have purchased 'previously owned' rigs, so it has been for me 'what you get is what you get'! But then again, I have not purchased rigs that I thought (general personal feeling - nothing concrete) were underpowered.

After reading your post, and relating it to my younger years (more years ago than I care to admit) and my previous involvement with hotrods and trucks - I ask:
When will we see the MOHO DRAGS?? That would be something - I wonder how my 'old' 460 pulling my rig would do against a much bigger and heavier newer one. I would love to watch, but I don't think I'll be doing it!!

1997 Rexhall Rolls Air-mid-entry DP w/cummins c8.3
Look for the Mardi Gras 'Gators - Laissez le bon temps roulez - Let the good times roll!! Best view of our rig-That's ME! Going Away!!

Johnny T Great answer on past topics. Another variable is the size of the engine generating the torque and horsepower. I can't even guess why this works the way it does but let me give you an example and perhaps others can explain. My old rig that I just sold had a Cummins ISM 11 liter engine. It produced 500h hps and 1450 ft/lbs of torque at 1200 rpm. My new coach has a Cummins ISX which is 937 cubic inches almost 14 liters. It produces 525hp 1650 ft/lbs at 1200rpm but does this seemingly effortlessly when compared to the ISM. The ISM is a great engine large by most standards but 20%+ less in size than the ISX. So why does displacement have such a great impact on torque and performance? Johnny if you read this thread please answer. Elliot

"There are many on the forum that already know more about these issues than I ever will. This post is aimed at those that are at the beginning of the learning curve. The example engines I used were the ones I have some experience with there may well be better examples. Hopefully those in the know will help the rest of us with some practical insights, that will aid those embedded in the selection process."

Torque is expressed in terms of ft pounds (sometimes expressed as pound ft.)The amount of force (pounds) developed is directly porportional to the diameter of the pistion, which also is a componet of "displacement" Displacement is the area of the piston times the length of the piston stroke. The more displacement you have the greater the force in the ft pounds calculation. This accounts for the old saying that "there is no substitute for cubic inches" Cubic inches equals displacement.

I found the timing of acceleration from 0 to 45 MPH and the then 45 to 60 to be a useful and recordable indication of performance.

JohnnyT

Not to quibble, but for some reason, magazines and power pack advertisements seem to advertise 40-60mph. Not much difference from 45-60, but handy to compare what you drive to what you read.

NOTE: Any incorrect spelling is intentional to prevent those annoying popups.

84 Barth 30Tag powered by HT502/Thorley/Weiand etc, Gear Vendors OD.
Siamese Calvin and Airedale Hobbes, 4WD Toyota toad

Brake Mean effective pressure!!! The horsepower generated by an internal combustion engine is determined by the volumetric efficiency of the cubic inches of displacement of the cylinders. In other words, how much of a charge you pack into a given swept volume of the cylinder by the piston and how much you compress that charge into the combustion cycle. The amount of BMEP changes with throttle opening and the general condition of the engine (an air pump analogy).
The better the compression the better the BMEP (condition of the piston rings cylinder walls and valves and valve timing).

The result is brake mean effective pressure which is then by formula converted to torque.
BMEP is the true indicator of power and takes into effect a frictionless machine. In other words if you had a BMEP meter in the cocckpit you would be looking at true output of the engine cause it's the net output.

The old round reciprocating airplane engines had a ring with forty pistons fed with oil pressure surrounded by a helical gear that pressed the pistons and measured the forward thrust of the helix that computed the Brake Mean effective Pressure the engine was producing.
We then prior to take off as we accelerated down the runway monitored the BMEP to see if a safe takeoff was in the cards. Real horsepower being delivered to the prop reduction gearing.

A truck is no different instead we are delivering the power from the engine (BMEP)coverted to torque (torsional rotation of the crankshaft) to the transmission via gear ratios to the rear end and finally torque delivered to the rear wheels and power being deliverd to the roadway taking into account the coefficient of friction between the tires and the roadway. The torque curves the manufacturers show are highly optimistic and are dynamotor curves that are just an indication of relative capacity of the machine.
A true indicaation would be considering dew point temp, outside air temp, barometric pressure (altitude) and show the relative torques available at altitudes we usually tavel at (like the rocky mountains)

* This post was
edited 03/15/06 08:31pm by angelino *