RCMAN46 wrote:
4x4ord wrote:
4x4ord wrote:
RCMAN46 wrote:
4x4ord wrote:
RCMAN46 wrote:
4x4ord wrote:
DougE wrote:
4x4ord - I think you just proved my statement on page 3 that torque determines how steep a hill you can climb and horsepower determines how fast you can do it.
Torque on its own is actually quite meaningless, but you are right that if two similarly geared vehicles were to pull similar trailers up an increasingly steep hill the one capable of producing more torque would be able to climb the grade further than the other. As well if the vehicles are geared such that they can best utilize their power the one making the most power will climb the hill faster than the other. The thing about all this is that engineers need to understand how HP Torque and gearing all relate and they design their vehicles to do the intended job. Marketing persons know that advertised torque and power numbers sell vehicles so they push the idea that more torque is always better or more power is better depending on what they are trying to sell.
Sorry we still have it wrong.
First an engine that can produce 400 ft lbs of torque at 1500 rpm.
Second an engine that can produce 300 ft lbs of torque at 5000 rpm.
You made the assumption we would gear each to utilize the power.
I also assumed you were talking about engine shaft torque.
The 300 ft lb engine will leave the 400 ft lb in its dust regardless of the load or how steep the hill is period.
As for advertising the Dodge Ram have for years tried to convince us that the engine that has the most shaft torque has the most power which is not true. But they have the Ram guys convinced that it is.
I actually don't have any idea what you're talking about. You are talking torque at rpm so you're comparing 286 HP to 114 HP. What do you think I have wrong?
"the one capable of producing more torque would be able to climb the grade further than the other"
If you had said (the one capable of producing more horsepower would be able to climb the grade further than the other) I would not have had a problem with what you said.
As you have stated before torque does not tell you anything about the power an engine can produce. You need rpm and torque and when you do that you have defined horsepower.
I agree with everything you have said with the exception of your quote I gave above.
Rather than re-explain things I will just try to state what I said a little differently: Consider to very similar trucks:
Truck #1 has a 6R140 transmission with a 3.73 rear end and a 400 HP Powerstroke diesel capable of making 800 lb ft of torque at the crankshaft.
Truck #2 is also a Ford F350 with a 6R140 transmission but has an aftermarket engine in it that makes 1000 HP @ 11000 rpm and a maximum torque of 600 lb ft @ 5500 rpm.
The two trucks are hooked to trailers and sent up a hill that keeps getting progressively steeper as they climb. Truck # 1 will climb much further up the hill than will truck #2.
If you still struggle with what I am saying can you explain why?
Edit:
Ive got a little time to add to this post to make things clearer. The gearing is responsible for multiplying the crankshaft torque. So both trucks have 3.73 rear axles and 3.97 first gear ratios. This means that while truck #1 is capable of delivering 400 hp to the rear axle the maximum torque it can deliver is 3.97x3.73 x 800 = 11,846 lb ft.
While Truck #2 is capable of delivering 1000 HP to the rear axle it is only capable of delivering 8,884 lb ft of torque (3,97 x 3,73 x 600).
If the tires have a diameter of 32" then the force the two trucks can exert against the asphalt would be 12/16 x 11,846 for truck one = 8,884 lbs of force for Truck #1
and 6,663 lbs of fore for Truck #2
Using vectors you can calculate that the necessary slope of the hill to put a pull of 8,884 on the outside of perimeter of the truck #1 rear tire's. Truck #1 would stall out once the hill got steeper than 25.5%
Truck #2 would be stalled out once the hill got steeper than 18.8%
Because there is no speed or rate of work in any of these calculations HP never enters the equations.
First no engineer would mate the 1000 hp engine you describe to the 6R140 which has a rpm limit of about 6000 rpm. To do it right he would first use at least a 2/1 planetary gear arrangement to get into the limits of the 6R140. But even then I suspect the 6R140 is not capable of handling 1200 ft lbs of torque. That would give about 35,500 ft lbs of torque at the wheels for the 1000 hp engine.
Now think about how I arrived at that ridiculous torque number.
I seriously doubt there is a set of tires that can even put down the 11,800 ft lbs on a 25% grade so neither truck would stall they would just start spinning tires.
None of this is anywhere near actual application of a towing truck.
I am sure you would have to agree if a F350 was ever to have a 1000 hp engine the engineers would mate it to a transmission and rear end system that could utilize the 11,000 rpm 1000 hp engine to its full potential.
I agree no engineer would couple an engine that revs to 11000 rpm to the 6R140 with a 3.73 rear end and put it in a towing truck. I used an exagerated illustration to make the point that if gearing and weighting is the same the engine able to make the higher torque will be able to climb a steeper hill....100% of the time even if the other truck has 2 1/2 times the power. You seem to have missed my point.
I have hooked my truck to 30,000 lb trailers and gone into soft ground where I ran out of torque but still had traction in 4 wheel drive. By dropping into 4 low it makes an incredible difference in those situations.
