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Automotive Horsepower and torque question

  1. May 22, 2015 #1
    Hi all nice site just trying to figure something out about how low RPM and high torque vs high RPM low torque for moving a load.

    This is a subject I have been trying to understand and it seems I am missing something, Ok take you have a 238 HP Mack Semi that makes 1100 ft pounds of torque and you have a 238 HP motor that spins like 6000 RPM to but does not make near the torque.

    Ok the Mack can pull 60 thousand pounds and has a top speed of only 67 MPH which is the 238 HP part and it only need s a 5 speed to do this.

    With proper gearing can the 238 hp 6000 RPM gas engine ever move the load like the Mack diesel or is the only thing they have in common is they will both run the same MPH but the Mack can haul way more weight to that MPH because it produces much more torque which is your ablity to build and maintain power under a load.

    This is only part of the ?, thanks and hopefully someone can get me straight on this.
     
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  3. May 22, 2015 #2

    SteamKing

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    You should understand how torque, RPM, and HP are related.

    HP = Torque * RPM / 5252

    The engine in the Mack truck is not turning at 6000 RPM like the car engine. For a torque of 1100 ft-lb and a power output of 238 HP, the Mack engine is turning about 1136 RPM. For the car engine turning at a speed of 6000 RPM making the same HP, the torque would be about 208 ft-lb, which is quite a bit less that the 1100 lb-ft from the truck engine.

    Usually, high revving engines like the one in the car have a speed where the amount of torque generated peaks and then drops off with increasing RPM. A truck engine is designed to operate at relatively low RPM, where the torque output remains relatively constant regardless of RPM.
     
  4. May 22, 2015 #3
    Ok I understand how to convert the Hp to Torque RPM just lazy, Thanks SteamKing. I am really realizing how dumb I am about math and how to do some figuring but it is never to late to learn. Here is the ?,

    You have 2 2300 HP motor units, one is a Pro Mod Drag motor that turns 9000 RPM and the other is a ship engine that turns 270 RPM but makes 41000 ft pounds of torque. Now the 2350 pound Pro Mod car will run 5.8 seconds and roughly 250 MPH in a 1/4. If you were to put the ship engine at the end of the 1/4 mile track and use a cable to pull the 2350 pound car with the right size pulley mounted on the ship engine and had it set up to where it is spinning it's max 270 RPM and then had a lock up clutch on it to where it locks down instantly would not the ship engine pulled car reach terminal velocity of 350 MPH in like an instant and run a sub 3 second 1/4 ?

    I had it figured to where about a 35 FT pulley on the ship engine would create the best time and go about 350 MPH which is about the speed a 2300 hp Pro Mod would achieve wound out in like a salt flat..

    Thanks for the help.
     
  5. May 22, 2015 #4

    SteamKing

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    You're still confused. The dragster can't go 350 MPH unless its wheels are spinning fast enough.

    Sure, the ship engine makes a lot more torque, but this torque is delivered at a very low speed (BTW, the torque from the ship's engine must be turned into thrust by the propeller in order for the ship to move at all.)

    If you were to put all 41000 ft-lbs of torque from the ship's engine into the dragster, all this additional torque would accomplish would be to burn up rubber from the tires. You want the tires to stick so that the car can move forward. You can calculate how fast the dragster would go by calculating the circumference of the tires multiplied by the RPM the tires are turning.

    A Top Fuel dragster tires are about 36" in diameter, which means the circumference of the tire is about 115". Assuming that the tire sticks, 115" x 270 RPM equates to a speed of 29 MPH, and that's assuming a final drive ratio of 1:1.

    The final drive ratio in the Top Fuel category is limited by rule to 3.2:1. For a dragster whose engine turns at 9000 RPM, the tires will turn at 9000 / 3.2 = 2812 RPM, which would equate to about 306 MPH, assuming no slippage between the tire and the track. That's as fast as the dragster can move, unless it starts flying (literally).
     
  6. May 22, 2015 #5

    billy_joule

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    If the two motors were set up in such a way that their power output remained constant (ie with ideal, continuously variable transmissions) then they would achieve the same 1/4 mile time (assuming ideal conditions - tires never slip, cables never stretch, no transmission losses etc etc)

    If the power of the two engines is the same they can achieve the same torque at the same rpm with appropriate gearing. That is, if you gear the boat motor up to output the same RPM as the dragster motor their torque outputs will be identical (or vice versa).
     
  7. May 22, 2015 #6
    This is a debate I am having in another forum and they are saying HP is HP and you just gear how you want but the people who own the diesel PU's that are rated at the same HP or less as their old gasser say no matter how they gear the gas the diesel just kills and I mean kills the gasser pulling heavy loads being able to pull fast under heavy loads and build MPH on steep hills..

    I am trying to figure this problem out, I talked to the guy's who build the fastest Pro Mod car engines in the world and they told me the ship motor 2300 HP 41000 LBS torque hooked to a cable pulling 2300 pound car would kill the 2300 pound car that has the 2300 HP car motor in reaching terminal velocity which would be about 350 for 2300 HP. They did not get into details about how who when where what and why however.

    They said the ship motor should hit 350 MPH in a snap and that would equate to a sub 3 second run. Maybe he was wrong , I am just trying to understand this problem and explain it in a simple way and this is what I came up with so please if anybody can show the math to explain it out better that would be great. I chose this comparison because they are total opposites of the piston motor world but have the same power rating.

    Here is the #'s I came up with on the ship motor for pulling 2300 pounds with a cable, please correct if I am wrong but please be right. The ship makes 41000 pounds of foot torque at 270 RPM, if you put a 35 FT pulley on this engine I worked it out you should get about if you pulled a cable with no load at 270 RPM it will travel 1/4 mile in about 2,7 seconds. Then you take 41000 ft pounds the crank is putting out then divide it buy 17.5 for the expanded 35 foot pulley conversion and I get 2343 ft pounds of torque on the outside of the 35ft pulley moving at a 1/4 mile every 2.7 seconds.

    This would lift the 2300 pound car straight up at 350 MPH because that is a 1/4 mile in 2.7 seconds ? The car motor car has the HP to run max 350 MPH but only has the torque to muster a 5.8 second 1/4 mile because the lack of torque cannot keep the RPM's pegged to maintain the 2300 HP constant from a standing start to the end of the track. It has a 3 speed and the RPM's drop

    In a horizontal line if you dropped a clutch down on the 35 foot pulley spinning at 270 RPM with 2343 pounds of torque on the cable on the outside of the 35 Foot wheel and locked it up the 2300 hundred pounds being pulled on wheels should run a sub 3 seconds and reach 350 in a snap ? Thanks for any help
     
  8. May 22, 2015 #7

    rcgldr

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    Assume that a vehicle with diesel engine and another vehicle with gas engine are both geared so that at the rpm of peak power for both vehicles, both vehicles move at the same speed. In this case, the torque at the rear wheels is the same for both vehicles at that speed. Note that power equals force time speed. So regardless of the torque, for a given speed, the force applied to the tires = (power at that speed) / speed. When geared to produce the same speed at the same power, gearing multiples the torque by a higher amount for the higher rpm engine, lower for a lower rpm engine, and the net result is the same rear wheel torque.

    The advantage of low rpm high torque engines is that they don't have to rev up very much to reach peak torque. There's also less wear on the engine when running at lower rpms.

    Note that Audi used high rpm diesel engines in their R10 Lemans race car (part of this was due to the rules for diesel versus gas engines), wiki article:

    http://en.wikipedia.org/wiki/Audi_R10_TDI
     
  9. May 22, 2015 #8

    billy_joule

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    Because the drag car motor is rated for X power doesn't mean it can apply X power to the road at all times. What happens when a drag car uses too much throttle off the line? So what happens instead? They use some fraction of full throttle off the line, right? So how much power is that? It's generally determined by the friction coefficient of the tyres.

    Does the winch have a fundamental limit on how much power it can apply? The strength & elasticity of the cable? The friction coefficient of the clutch? Do you think these are comparable to the drag cars fundamental limit?

    Power is the rate of doing work. Forget torque & RPM. Power is what gets work done.

    http://en.wikipedia.org/wiki/Power_(physics)

    Power = Energy / time
    P = E/t
    In SI units: Watts = joules / seconds


    http://en.wikipedia.org/wiki/Kinetic_energy

    Energy = 1/2 * mass (kg) * velocity (m/s) squared.

    E=1/2 m v^2

    So to get a 2350 lb car to 350mph you need to apply a given quantity of energy.

    http://www.wolframalpha.com/input/?i=1/2*2350lb+*+350mph^2

    36.94 kilojoules

    The power determines how long it takes to supply that energy:

    P = E/t
    power = energy / time

    t = E/P = 36.94kJ / 2200HP = 0.02 seconds

    http://www.wolframalpha.com/input/?i=36.93kJ+/+2200HP

    (I've used wolfram because it converts units where required automatically)

    The time is tiny - 2/100ths of a second. We don't see this in practice because of losses & limits to power applied (ie power curve shape, aero drag, tyre friction, transmission losses etc etc etc)


    No, that is all wrong. Your analysis ignores the effect of the load. That is, you assume your which will accelerate a matchbox car at the same rate as it would to the empire state building.
    Your winch cable will apply a force, the acceleration of the load will depend on mass of the load via
    F = ma
     
  10. May 22, 2015 #9

    rcgldr

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    The engines for the top dragsters (alcohol, nitro-methane), are launched at full throttle and high power output, and the clutch is mechanically programmed to consume power (conversion to heat) by slipping at launch and then gradually engaging to reduce power consumption during a run until the tires can handle the torque related to power at some speed (full engagement and no clutch slip). There are clutch fingers that use centrifugal force to apply pressure to the plates, combined with some type of pneumatic timers (air, co2, oil, ..., electronic timers are not allowed) that controls the rate at which the clutch fingers are allowed to increase pressure to the clutch plates.

    The clutch programming is a key aspect of drag racing, too conservative and the run is slower than the competition, too aggressive and the tires spin. The rules for drag racing in these classes do not allow electronic based traction control, but I don't know if the rules allow for rev limiters to limit rpms during the early part of a run when the clutch is slipping a lot.

    For other classes like pro stock cars and pro stock motorcycles, there are two rev-limiter settings, one while staged just before launch, and then max rpm setting for the actual run, and the driver controls the clutch. The clutches may be setup to slip a bit for the initial launch (I'm not sure about this).

    Youtube video on top fuel clutches:

     
    Last edited: May 22, 2015
  11. May 22, 2015 #10

    billy_joule

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    Interesting. Is the driver allowed to steer the car? Or is that taken care of too? :biggrin:
     
  12. May 22, 2015 #11

    rcgldr

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    There's a staging brake that holds the car in place once it's staged (the front tires blocking the second staging beam of light), so the driver just hits the throttle (for some classes of drag racing, a switch to disengage the staging brake is used), and hopes to keep the car pointed forward with steering inputs. For top drag classes, the drag race starter flips a switch to turn on the yellow lights (all of them at once), and the driver has to react quickly, but not too soon so that the front tires do not clear the second staging beam (about 12 to 18 inches traveled) before 4/10ths of a second for the top classes, and 5/10th's of a second for lower classes, otherwise it's a red light (disqualification). For some lower class events, the yellow lights turn on sequentially, so the driver is mostly timing the sequence of lights rather than quickly reacting to lights just turned on all at once.
     
  13. May 22, 2015 #12

    SteamKing

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    If you make a big enough wheel, you can do a quarter mile in less than a second. However, it takes a lot more torque to get a 35-ft diameter wheel turning than a couple of wheels which are only 36" in diameter.
     
  14. May 23, 2015 #13
    Thanks for all the help I will look it over so in the future I understand this on a math perspective.

    Basically set me straight here. I know about the clutch and tire spin on a fuel car. A 2300 HP Pro Mod on a good track can just drop the clutch solid and will not spin the tires hardly at all and needs a 3 speed transmission to get desired results of 5.8 seconds at around 250 MPH.

    If the Pro Mod car were geared to where it would max out at 350 MPH which is about 2300 HP and you did a 1 gear ratio that was that peak MPH with no shifting and dropped the clutch on the starting line it would stall instantly if the tires did not spin to allow the engine to regain RPM to unload it ? this is just lets say.

    If I calculated correctly which I probably did not possibly, my math needs work. If you had a force that is moving at 350 MPH with 2400 torque pounds of pull and had it push a 2300 pound wheeled sled from a standing start how long would it take the 2300 pound sled on wheels to reach 350 MPH ?
     
  15. May 23, 2015 #14
    The reason I am asking this is because I am trying to figure a way to explain this to people to where they understand it without getting into the math of why a motor that makes more torque than another can pull and reach terminal velocity faster than a motor which is rated the same HP but makes less torque because it spins more RPM to make the HP.

    I could very well be wrong about this but the people who build the Fastest Pro Mod engines in the world told a ship motor geared right with 41000 foot pounds of torque and 2300 hundred HP pulling would reach terminal velocity way faster than a V8 car motor spinning 9 grand to make 2300 HP and they said it was not tire spin but the ability of the ship motor to maintain RPM HP under a load because it possesses so much torque.

    I under stand what you all are saying about forget about RPM torque here and it is just power needed but ?
     
  16. May 23, 2015 #15
    Thanks Billy, I appreciate it alot. No you could put a Pro Mod Engine on a stand at the end of the track and do same thing hypothetically. Let me go over all this, I am as thick as a brick so try and have patience.
     
  17. May 23, 2015 #16

    rcgldr

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    Power = force x speed. If using a cable to tow the 2300 lb car with a 2300 hp engine with a continuously variable transmission that keeps the engine running at the rpm of peak power, and the situation is idealized so there are no losses, then force = power / speed = mass x acceleration, so acceleration = power / (mass x speed) as speed goes from zero to the speed after some amount of time or distance. The initial instantaneous acceleration is infinite, but the speed or distance versus time or time to speed or distance are finite values, and the same regardless of the torque versus rpm, since the transmission multiplies the torque by a greater amount for the higher rpm engine and a lower amount for the lower power engine. Note that the following formulas are based on power, not torque:

    [tex] p = f \ v [/tex]
    [tex] a = \frac {p} {m\ v} [/tex]
    [tex] v = \sqrt {\frac{2\ p\ t}{m}} [/tex]
    [tex]t = \frac{v^2 \ m} {2 \ p}[/tex]
    [tex] x = \sqrt {\frac{8\ p\ t^3}{9\ m}} [/tex]
    [tex] t = \sqrt[3] {\frac{9\ m\ x^2}{8\ p}} [/tex]
    [tex] v = \sqrt[3] {\frac{3\ p\ x}{m}} [/tex]
     
    Last edited: May 23, 2015
  18. May 23, 2015 #17

    SteamKing

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    You're getting into things where the real world has a say.

    The ship engine which makes 2300 BHP @ 270 RPM is physically a bigger and heavier mechanism than an automotive V8 which makes 2300 BHP @ 9000 RPM.

    For example, a MAN 8L21/31 marine diesel engine produces 1720 KW @ 1000 RPM, or about 2305 BHP, but this unit weighs 19,000 kg (41,900 lbs) dry and is over 17 feet long. An engine of similar output at 270 RPM will be even bigger and heavier than this MAN unit. By contrast, the V8 engine probably weighs 500-600 lbs w/o transmission and is maybe 3 feet long.

    The moral of this story is you can't cherry pick one set of performance specs from one type of engine and expect them to relate to different applications entirely. You wouldn't want to build a dragster using a ship engine any more than you would want to build a ship using a drag engine, even though both engines are capable of putting out the same power. Among other things, the ship engine is designed to run for days at a time and last for years, while the drag engine is capable of running flat out only for a matter of seconds.
     
  19. May 23, 2015 #18
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    Last edited: May 23, 2015
  20. May 23, 2015 #19
    I am still lost so please bare with me, I cannot read the formulas above but need to learn. Thanks for taking time.

    Some of the inputs I used were wrong so I will try again. So HP is HP and you gear right it is the same ?

    The motor that is 2300 HP which is all that matters you all say the rest is window dressing of use. It makes 44735 FT pounds of torque at 270 RPM. You put a 35 foot wheel on the end of it and it will pull a cable with no load 1320 FT every 2.66 seconds. How much torque would be on the end of the cable being pulled and how fast would it pull a 2300 pound sled on wheels in 1320 feet. I am talking with the motor running 270 RPM and you drop a clutch on the 35 foot pulley. This is all hypothetical, forget the little details just a rough guess using proper math.

    I appreciate the time you all have taken.
     
  21. May 23, 2015 #20

    rcgldr

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    There would be a greater reduction of rpm in the gears for the higher rpm motor.

    Say the ship motor makes 2300HP at 270 rpm, the drag race motor makes 2300HP at 8910 (33 x 270). The ship engine makes 44740 fl lbs of torque at 270 rpm, the drag race motor makes 1355.7575... ft lbs of torque at 8910 rpm, but say it's initial gear reduction stage is 33 times that of the ship engine, so this multiplies the torque of 1355.7575... by 33 = 44740 ft lbs at 270 rpm after the gearing.

    In the real world, the low rpm, high torque engine would be harder to lug and get stuck below it's peak power rpm, but there are gas engines with nearly flat torque curves that could be used.
     
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