Math Question

Ok you math gurus.

I have a question on torque and hp. If I have a transmission that will produce 100 hp at 2000 rpms and 1400 ft lbs of torque, what will the torque and hp be if I make a 1 to 2 ration and increase the rpms to 4000?

Thanks

Mark

 

YOU HAVE not made it clear if you are increasing engine revs or if your gear ratio is an up rather then down gearing ..you will still have 100hp whatever you do but an up ratio of 2:1 will give you 700 ft lb ( gearing efficiencies neglected )

If you mean inc engine revs to 4000 and then put a 2:1 reduction then you would need to know your engine torque at 4000 rpm and double it

 

I was talking about using a 1 to 2 gears ratio. I am starting with a motor that turns 2000 rpm, produces 1400 ft lbs of troque at 2000 rpms. the rated hp is 100 hp. now attach a 2 inch pulley to the motor, and a belt that goes to a 4 inch pulley which increases the rpms to 4000. this would drive the shaft to turn the prop. What effects does this have on torque and hp?

 

Mark,

The relationship between torque, rpm, and horsepower is:

hp = (torque * rpm) / 5252 where torque is in ft*lb. That means a 100 hp engine will produce 263 ft*lb of torque at 2000 rpm.

If you double the output rpm of your engine, your output torque will be half of what it was before. Horsepower does not change.

 

if you have 2 inch pulley on motor and 4 inch on shaft you will halve your rpm and double the torque BUT how can you transmitt 100 hp through a 2 inch pulley .....no you cannot ..maybe 4 hp ...you need a gearbox not a pulley system
1400 ft lb is far too much for a 100hp (diesel ) maybe 300 ......mmm new engine type maybe .....

mind you he does say a transmission that will produce 100hp .....transmissions produce nothing they absorb power ..
who is the confused one here????

 

1400 lbs of torque at 2000RPM indicate about 400HP. However if you double your speed it will half your torque, minus mechanical losses.

 

Thank you all for the help,(except you pisnbroke) I am trying to learn something here about Hydraulics and relation to IC engines. So to the rest of you who like to help and not be a a$$ hole, here is what I am trying to figure out. If I have a 100 hp IC engine and I attach a Hyd pump to it, and that creates a flow rate of 33gpm at 1800 rpm. Now the engine turns optium at 4000 rpm. So I think I need to reduce the gear ratio to 2 to 1 to get down to the max application of the Hyd. pump of 2000 rpm. Now I attach a Hyd motor to the pump system and it will run at 33 gpm and produces 2000 rpm. The data says the Hyd. motor will produce 1400 ft lbs of torque. But I want to run a smaller prop, so I want to increase the rpms to accomplish this. What I am thing is to reverse the gear ratio back to 1 to 2 past the Hyd. motor to create 4000 rpm. This is just a thought and that is why I asked this stupid question. I wanted to know the effects of changing the ratio back to 1 to 2 and increasing the rpm created by the Hyd, motor.
Your thoughts? (pisnbroke you can keep your thoughts to yourself).

thanks guys

Mark

 

Don't take it personally - Australians, known for their educations, are actually one of the rare demographics qualified to dis our system (and he's a good guy. First time I've seen this from him.).
You see, what many people don't grasp is how much of a melting pot we are, nor how difficult it is maintaining such a bastion of freedom as the U.S., while dropping ladders for the persecuted from the four corners. Instead of shipping off our undesirables to far-off God-forsaken ****-hole continents, we try to meld them into our fabric and make useful citizens of them. Anyone can leave here at any time, and with a modicum of effort, can join us to defend this last stand of freedom on Earth - and we are being overwhelmed as an immigrant nation.
Expect to hear open forum people using exported American technology, freedom, and generosity to try to offend the very place that has so often and so readily given to defend their freedoms. Expect to be called a "septic" (one of their favorites), and just let it roll off your back like I do but DO remember how they are when they one day, once again, get their **** in a ringer. Have a good day.

 

Thanks Mark for the info. However, most of the post that I have made on here....pisnbroke has been in my face. Just wanted to set him straight as to quit being a smart A$$ and don'toffer help if I bother him. If he ask me questions on my expertice, (Aviation) I would be glad to help him out no matter what the question was. I had an instructor tell me years ago that no question is dump until you ask it the tenth time...lol Anyway this is getting off thread and I still need help in understanding the basics of Hyd.

regards

Mark

 

A hydraulic motor will have a loss of about 30%, maybe more if it is larger than needed

 

Aside generalizations about hydraulics there's a huge difference btw open and closed circuit hydraulics. With closed circuit effiency about 80 to 90 is quite reasonable to expect and is the choice for continous duty.
What comes to your question it's best to choose pump to meet your engines rpm/power output directly like this http://www.boschrexroth.com/mobile-...?Language=EN&VHist=g54069,g55969&PageID=m3544 the biggest one in this page with it's rated rpm being 3000..

 

The thing to look at is probably the weight of the boat vs the power you have. If the boat is relative easy propelled, you can up the speed and get away with lower torque.

The opposite also applies, if the torque is not enough you can reduce speed and increase torque.

If the drive combo is good (near optimal), you won't get more speed by doubling the ratio ! The motor will labour and you'll be heavier on fuel.

Pulley's are not very effective, all mechanical contraptions have losses. The sprocket belts and 'timing' pulleys you get nowadays are probably the best if you go that way, visit the nearest bike shop to check them out.

Most adjustments of power to speed and torque is usually done by changing the prop ratio only, going double on a drive is quite radical.

I think pistnbroke (pisnbroke ) made a joke. Lighten up or we'll have to put a reduction pulley in your temper as well

 

Just curius: why this question.
Related to what?

 

dsk,
My thought was wanting to know how much powewr loss I would create by increasing the speed through a 1 to 2 pulley, (i do know about gears and timing belts pulley is just a way to say the ratio increase) Like riding a bike with many gears, you can go faster with a 1 to 2 ratio, but it takes more energy to create it. So I was trying to see what the rate of increase torque is needed to increase rpm. Does that make sense?

Mark

 

exactly dskira .....what do we have

a 100hp 1400ft lb engine ..not possible what type is it ?? Oh its hydraulic so is that a stall torque /continuous or intermittant max...you will need way over 100hp to drive the pump....
Its driving via a 2 inch pulley .....not possible .....

fortunaltly I dont do hydraulics so go for it and I will watch...