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dumb question about trq.

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    Resident Ford Nut Sleeper CP's Avatar
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    Default dumb question about trq.

    lets say you have a low reving engine that makes 800 lbs ft. at 3,500 rpm. If you put a 2:1 gear box behind it and the out-put shaft is spinning 7,000 would that be the same as having 1,066 hp ? ( less any loss to spin the gear-box)

    S CP

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    gn7
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    Quote Originally Posted by Sleeper CP View Post
    lets say you have a low reving engine that makes 800 lbs ft. at 3,500 rpm. If you put a 2:1 gear box behind it and the out-put shaft is spinning 7,000 would that be the same as having 1,066 hp ? ( less any loss to spin the gear-box)

    S CP
    Nope. any time you gear uup you gain RPM but lose torque, same goes the other way. Its why you car pulls like hell in low gear. Boat loads of torque traded for
    RPM. My V-drive has less TORQUE at the prop than at the crank because of the overdrive. But it stills has the same HP minus frictional loss from the drive train. You don't lose HP(again except frictional losses) fro the gearing, you lose torque and gain RPM. Reverse it, and you gain torque and lose RPM. At the end, the Torque X RPM / 5252 still hold true. Make sense?



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    Highaboosta Unchained's Avatar
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    That's a good explanation Bob.
    I understand about torque multiplication and torque division but I never considered that the HP would remain the same, it just changed RPM.

    Twin Turbo 1800 HP V-Drive lake boat

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    Quote Originally Posted by Trigger View Post
    No one cares about your buddies old antiquated garden hose technology.
    Quote Originally Posted by MAXIMUS View Post
    I think I could run more boost but it's a real hand full right now

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    Resident Ford Nut Sleeper CP's Avatar
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    Quote Originally Posted by gn7 View Post
    Nope. any time you gear uup you gain RPM but lose torque, same goes the other way. Its why you car pulls like hell in low gear. Boat loads of torque traded for
    RPM. My V-drive has less TORQUE at the prop than at the crank because of the overdrive. But it stills has the same HP minus frictional loss from the drive train. You don't lose HP(again except frictional losses) fro the gearing, you lose torque and gain RPM. Reverse it, and you gain torque and lose RPM. At the end, the Torque X RPM / 5252 still hold true. Make sense?

    No............ maybe bp can explain it to me ............

    Power plant is turning at it's max load/power 800 lbs ft at 3,500 if the gear box is spinning at 7,000 how am I losing power ?? other than the drive-line loss ?

    if I am losing power ( or trq) because of rpm how much would it be ?


    Something told me it wasn't straight across but I don't get it........at least not yet.

    Edit: So if the gear box takes 85 lbs ft to spin at 3,500 so the net at the out-put shaft would be 715 lbs ft that would be 953 hp vs 1,066 at 7,000?

    S CP
    Last edited by Sleeper CP; 06-29-2011 at 09:04 AM.

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    Senior Member bp298's Avatar
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    Quote Originally Posted by Sleeper CP View Post
    No............ maybe bp can explain it to me ............

    Power plant is turning at it's max load/power 800 lbs ft at 3,500 if the gear box is spinning at 7,000 how am I losing power ?? other than the drive-line loss ?

    if I am losing power ( or trq) because of rpm how much would it be ?


    Something told me it wasn't straight across but I don't get it........at least not yet.

    Edit: So if the gear box takes 85 lbs ft to spin at 3,500 so the net at the out-put shaft would be 715 lbs ft that would be 953 hp vs 1,066 at 7,000?

    S CP
    sorry jon, can't explain it any better than gn did.

    your "edit" example is confusing. i could guess what you're asking, but not gonna do that. i do not understand your questions about "losing" power.

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    Highaboosta Unchained's Avatar
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    Look at it this way,
    Say you have an I/O which is 2:1 gear ratio underdriven.
    5000 rpm and 500 ft lb input is only 2500 rpm output rpm but torque is multiplied by 2 to 1000 ft lbs.
    The input gear has more leverage against the output gear.

    Now consider a vdrive with 2:1 overdriven gears.
    5000 rpm and 500 ft lb input gives 10000 rpm output but torque is divided by 50% to 250 ft lbs.

    Twin Turbo 1800 HP V-Drive lake boat

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    Quote Originally Posted by Trigger View Post
    No one cares about your buddies old antiquated garden hose technology.
    Quote Originally Posted by MAXIMUS View Post
    I think I could run more boost but it's a real hand full right now

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    Senior Member bp298's Avatar
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    Quote Originally Posted by Unchained View Post
    Look at it this way,
    Say you have an I/O which is 2:1 gear ratio underdriven.
    5000 rpm and 500 ft lb input is only 2500 rpm output rpm but torque is multiplied by 2 to 1000 ft lbs.
    The input gear has more leverage against the output gear.

    Now consider a vdrive with 2:1 overdriven gears.
    5000 rpm and 500 ft lb input gives 10000 rpm output but torque is divided by 50% to 250 ft lbs.
    ... with no frictional losses...

  10. #8
    cfm
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    Gears multiply power inputed to them by their ratio.
    Gears divide rpm inputed to them by their ratio.


    EX:
    400ftlbs from engine into a 4:00 gear ratio = output of 1600ft/lbs
    5000rpm from engine into a 4:00 gear ratio = output of 1250 rpm


    Hope that makes it easier.

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    Highaboosta Unchained's Avatar
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    Quote Originally Posted by bp298 View Post
    ... with no frictional losses...
    Well I certainly don't have any frictional losses with my setup ........

    Twin Turbo 1800 HP V-Drive lake boat

    http://s621.photobucket.com/albums/t...t=MAH05771.mp4

    Quote Originally Posted by Trigger View Post
    No one cares about your buddies old antiquated garden hose technology.
    Quote Originally Posted by MAXIMUS View Post
    I think I could run more boost but it's a real hand full right now

  12. #10
    cfm
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    Quote Originally Posted by Sleeper CP View Post
    No............ maybe bp can explain it to me ............

    Power plant is turning at it's max load/power 800 lbs ft at 3,500 if the gear box is spinning at 7,000 how am I losing power ?? other than the drive-line loss ?

    S CP
    Forget about frictional hp for a minute to a grab a hold of what is going on.

    Your engine is not losing power. It is what it is. However, actual shaft power, with your gear ratio example, the power after the gears will be less than the motors.

    Gear Ratio = Power Multiplier (and output shaft rpm)
    2:1 gear ratio = 2 (shaft rpm will be 1/2 the engine's rpm)
    1:1 gear ratio = 1 (shaft rpm will equal the engine's rpm)
    1:2 gear ratio = 1/2 (shaft rpm will be double the engine's rpm)

    Don't think about cars in respect to driveshaft rpm because the driveshaft is before the rear-end gears. This may trip a boat owner up.

    800ft/lbs at 3500 engine rpm
    800ft/lbs X 2 = 1600ft/lbs at 1750 output shaft rpm
    800 X 1 = 800ft/lbs at 3500 output shaft rpm
    800 x 1/2 = 400ft/lbs at 7000 output shaft rpm

    Don't confuse engine tq with output shaft torque. Don't confuse engine rpm with output shaft rpm. They will only be the same if a 1:1 gear ratio.

    So, the question if one is talking prop driven boats is, do I add more output shaft torque and run a bigger prop slower, or do I reduce output shaft torque and run a smaller prop faster ? The answer to this is definately application specific. Many variables. Type of drive system, boat weight/design/intended use, style of prop, engine, and etc.

  13. #11
    cfm
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    Quote Originally Posted by gn7 View Post
    Nope. any time you gear uup you gain RPM but lose torque, same goes the other way. Its why you car pulls like hell in low gear. Boat loads of torque traded for
    RPM. My V-drive has less TORQUE at the prop than at the crank because of the overdrive. But it stills has the same HP minus frictional loss from the drive train. You don't lose HP(again except frictional losses) fro the gearing, you lose torque and gain RPM. Reverse it, and you gain torque and lose RPM. At the end, the Torque X RPM / 5252 still hold true. Make sense?
    Are you saying the following about shaft horsepower?

    Since we know the following equal horsepower

    TqXRPM
    ------
    5252

    and using a 2:1 gear ratio for an example:

    (2 X TQ) X (1/2 X RPM)
    ----------------------
    5252

    Math says it's the same hp being (Tq X RPM) / 5252


    If so, I can say I really never looked at this part of it either but have followed those guidelines...if you know what I mean.
    Last edited by cfm; 06-29-2011 at 04:41 PM. Reason: ooops. typing fart.

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    Senior Member ol guy's Avatar
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    This is cool, but I an not getting caught in this trap again!!!! It boils down to a designed weight-mass-movement. I will say torque gets it moving and H/P gets it going. Then toss in gearing, prop, impeller and weight of what is being moved. M

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    Resident Ford Nut Sleeper CP's Avatar
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    I was just thinking out loud that if someone wanted to build an impeller dyno ....would you be better off using a trq monster diesel engine and then over-driving it to get to 7,000 rpm or would you build a blown or TT gasoline engine to make 1,100 -1,200 hp at 7,000- 7,200.

    Just a "junk yard wars" thought.

    S CP

    "Dark Sarcasm"
    Going fast is only half the fun ... what you make go
    fast is the other half.
    " A Government big enough to give you everything you want is big enough to take away everything you have"

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    Senior Member ol guy's Avatar
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    Quote Originally Posted by Sleeper CP View Post
    I was just thinking out loud that if someone wanted to build an impeller dyno ....would you be better off using a trq monster diesel engine and then over-driving it to get to 7,000 rpm or would you build a blown or TT gasoline engine to make 1,100 -1,200 hp at 7,000- 7,200.

    Just a "junk yard wars" thought.

    S CP
    Hey sleeper! It's actually a good topic and very disregarded ! I have gotten beat up over this before for different applications. Building a impeller dyno would be tough unless you had a computer to factor in every aspect of the designed useage- Weight-drag coeficient effect and intended design. Then you need to factor in the horsepower and torgue rateings -then toss in intended usage. Which I think you have done quite well with your'e G/W. You seem to be getting the most of both worlds and maybe giving up a touch on one or the other! It seems to me you have built a boat that meets your're needs and added a sh$t pile of money to get there! JMO M

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