# Math Question

Discussion in 'Propulsion' started by Mark Emaus, Mar 18, 2010.

1. ### dskiraPrevious Member

Certainly it make sense in a way, but what will be the finality if you get
I was just curius if you are thinking about a new boat, or a new design, you
know something like that. Size, weight, style......
It's always interresting to look at new stuff comming and different
approach.
Daniel

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### mreoe4surewho me

mark, I hope this helps. The reason sometimes things might be said that are not helpful, is we are trying to figure out what the question is about. We say to ourselves what is he trying to do, It doesn't sense, but I hope this will help. Paul gave you HP = (Torque x Rpm)/5252 that is a given for Interal combustion engine. You said the HYD pump puts out 33gpm at 1800 rpm, but what power is required to produce this output. That is were some of the questions come from. There is a torque curve and a Hp curve on an Ic engine and you said 4000 was the optimum. Optimum what Hp or torque? It is why some people might send you to sites to help with pump selections. You most likely would not run an engine at 4000 rpm full time or you would need to replace is often. What was said is the engines Hp does not change with gear change, the torque output (the twisting force) changes. Just like a bike your power output does not change, the torque you put out is changed by the different gears you choose, so if you 1/2 the rpm you double the torque, if you slow to 1/4 the rpms by gearing torque is 4 times, the inverse is true, if you speedup the rpms the torque will drop. If you look at the formula you will see how this works. Now to the other fact that are important, every time you go through one of these torque multipliers, or RPM increasers ( gearing, pulleys, hyd pump, hyd motor, friction loss through hoses and fittings) you are robbed of some of the power. You will see as Gonzo wrote there might be up to 30%, and he is pretty sharp. If you just look at some of the threads around this site you will see this talked about , cpp with Apex1, diesel electric, hybrid, etc. No question is dumb, sometimes we just don't understand what is being asked. Hope this helps Steve

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### alan whiteSenior Member

Mre is correct. Horsepower remains the same---- any speed increase reduces torque and any torque increase reduces speed, and the means by which the conversion takes place determines how much horsepower is lost in the form of heat (no speed conversion is 100% efficient). However, no horsepower is lost due to the actual speed increase or decrease.
Hydraulics, as said by Gonzo, are quite inefficient compared to other forms such as gears or belts. However, the flexibility of hydraulic drives is sometimes well worth the loss of power. Everything depends on what you're trying to achieve.
Overall efficiency considers, for one thing, keeping the engine running at an rpm that uses the least fuel to produce the most power regardless of output rpm requirements. This same principle is how the Subaru Justy with a variable CVT drive gets better mileage than the exact same car with gears (fixed ratios)--- even though the gears have lower frictional power loss. Letting the engine run at its happiest rpm often makes up for what would be significant power losses at fixed ratios.

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### Mark EmausJunior Member

Thanks guys for the help.
The reason for Hydraulics in the first place is just as some of you have said. I am not as concerned with fuel efficiency as with application. I am willing to give up 30 percent loss of power as long as the drive train can sustain a harsh environment. I got a chart from a hyd. manuafacture on hp. needed to drive a certain gpm output of a hyd. pump. Our ic engine falls within those guide lines however, I am trying to max out the most torque and rpm as possible in the fitting of the prop. The idea of resizing the final drive back to a 1 to 2 ratio and increasing the rpm back to 4000 rpm was to decrease the prop size to goin to shallower water. I am planing on the boat being able to go into 7 inches of water with my design and would like the prop to stay in the water at that depth. So going from a 15 in prop to a 12 prop may make a big difference. I have changed designs many times in getting this whole project to work, but that is what make it fun. Not to mention learning a whole power source to drive the boat. So trying to find the loss of power by going back to a higher gear ration on the final drive was the sole purpose of this thread. I know there will be no change in the relation of the ic engine to the hyd pump, (they are compatable). The concern is the hyd pumps relation to the hyd.motor to drive the prop, along with sizing the hyd motor in rpm and torque. So far I have a pump with 4.93 cir output of 38 gpm at 1800 engine rpm and a max of 1500 psi, I am trying to size a hyd motor to this pump. Right now I am looking at a vane motor that will turn 2000 rpm at 33 gpm creating 1450 in lb of torque. As you can see, this is why I am thinking of a 1 to 2 ratio back to the shaft to turn the prop at 4000 rpm andwondering what the loss of troque would be if I did this? WHEW! that was a lot. I hope that made sense...By the way, this is an open system.
Mark

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### TeddyDiverGollywobbler

Have thought about the possibility to have several hydr.motor/propeller units ? that would give you more total propeller area and lower rpm..
Open circuit is not very good in this. Remember cooling!

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### pistnbrokeI try

1400 inch pounds now not the foot pounds of post one ....still watching and learning ....

7. ### FrostyPrevious Member

The magic word is IF --he said "IF he had a" --he is offering easy numbers to help you to show him what the results would be.

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### pistnbrokeI try

come on frosty IF he says foot lbs and is an aero engineeer he means foot lb .....if he torqued bolts on his plane to 12 times the correct torque would it stay in the air ??? schoolboy mistake ..

but forget that let help him solve his problem

9. ### FrostyPrevious Member

Piston,-- your right he did get confused with that, but "you" panicked him.

He did'nt say he was an aero engineer, he said he was in aviation. he did'nt say he had a plane or that he torqued fastners.

Yes lets solve his question.

So 2000 increased to 4000 by a 2:1 ratio would result in half the torque twice the RPM and 100 HP remained 100HP ?

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### Mark EmausJunior Member

Common sense…..how refreshing! By the way Pis, I fly them, I don’t fix them.

Thank you Frosty

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### Mark EmausJunior Member

One more dumb question...If you do have a torque in inch pounds, If you want to know foot pounds, do you divide by 12? In this case 1400 inch pounds would equate to 116.6 foot pounds? Just want tomake sure the motor will handle the prop.

Mark

12. ### Paul KotzebuePrevious Member

Yes. 1400 in*lb * 1 ft/12 in = 116.67 ft*lb

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### Mark EmausJunior Member

Ok Jango and the prop sizing guys. If we are running this pontoon at 2,000lbs and we are turning a prop at 2000 rpm with 116 ft lbs of torque, what size prop and what speed can we expect. 2nd. if we now turn the prop at 4,000 rpm bygearing to a 1 to ratio and only have 58 ft. lbs of torque, which would be better performance, (if it makes a difference)? What would the prop size be for that power? These pontoons arenot round but rather more flat on the bottom with a slight V and are 22 inches wide.

Mark

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### pistnbrokeI try

Thanks frosty my good friend (ha ha ) ..but I would not right a question with such a gross error .
Regular posters will realize we do get "joke posts" and " do my PHD for me " questions so apologies to the original poster ..now those in the know have the facts you will get good advice .

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### Mark EmausJunior Member

No Problem Pistn. I also should not have come down on you so hard.

TC
Mark

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