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View Full Version : Horse Power, RPM, Torque?



Lowell Bernhardt
09-28-2002, 09:09 PM
I was looking at the Ken Hinkensin plans catalog yesterday. In the front of the catalog there is the "SLOT MACHINE". A 45'fiberglass screamin demon. The book says that you can put twin 1100HP I/O units on this beast. Claimed speed will be somewhere around 60MPH. Aside from it being built of, well, you know. I've got a question.

If you were to take 3 of the exact same boat, with the same prop, same weight, same everything. And just for the purposes of this experiment put 3 different motors in them. One motor has the highest HP rating. Another has the highest torque rating, and the last can be driven to a higher RPM than the other 2. All the motors, we'll assume that all the motors are very close in weight and performance, with the exception of the mentioned criteria.

OK, now to the point. Which boat will go faster? What do you want more of to make a boat fly, HP, RPM, torque?

Dan McCosh
09-28-2002, 10:37 PM
Problem with this comparison is that horsepower is a funciton of torque multiplied by rpm. Torque and rpm are interchangeable via a gearbox., "all things being equal" assumes the gearing will favor rpm or torque, but not both. The pitch of the propellor inevitable will be optimized for either a high rpm, low torque engine or a high torque, low rpm engine, but not both. Answer ends up being another question-- which engine best matches the prop size and pitch, and how is it geared. Assuming the prop is sized for a high horsepower engine, obviously this will be the fastest. (also, most of the time a high horsepower engine with have both high torque and rpm.)

John E Hardiman
09-29-2002, 11:10 AM
What Dan said :D
To drive a hull at a given speed(its velocity, V)requires a certian force (its drag, Fd). This force times speed in feet per second divided by 550(Fd*V/550)gives the effictive horsepower (ehp) required to drive the hull.
Now for a propeller to produce a given force (its thrust, T) at a given pitch (its P), diameter (its D), and speed through the water(its advance, A) requires a certian rpm(its n, normally given in rotations per second on this side of the pond) and a certian shaft torque (its Qs). Therefor the thrust horsepower (thp) generated by the prop is T*A/550 and shaft horsepower (shp) required to generate this thrust is 2*pi*n*Qs/550. The prop efficency, nu, is therefore (T*A)/(2*pi*n*Qs). The power required by the engine (the brake horsepower, bhp) is the shp times the gear box efficency, nug, (~0.9) and the reduction ratio,r, relates the engine rpm to the shaft rpm inversly proportional to torque. So the engine torque is Qe=Qs/nug*r and engine rps ne=rn and bhp=2*pi*ne*Qe/550=2*pi*n*Qs/(nug*550). Note that this is for a specific propeller P vs D, blade geometry and a specific combination of A and n.

Now advance (A) is related to boat speed (V) by the wake fraction, w, (A=V*(1-w) in Taylor notation or A=V/(1+wf) in Froude notation for those of you across the pond). Given the bounding equlibrium condition that ehp=thp then we can write:
Fd*V/550=V*(1-w)*T/550 > Fd=(1-w)*T
substituting for T with nu gives
Fd=(1-w)*2*pi*n*Qs*nu/A
substituting A/V for (1-w) gives
Fd=2*pi*nu*n*Qs/V
substitute ne for n and Qe for Qs we get
Fd=(2*pi*nu/V)*(ne/r)*(nug*r*Qe) >>
Fd=(2*pi*nug)*(nu/V)*ne*Qe

As can be seen the equation becomes indeterminate because the term (nu/V) must be re-determined for each change in ne or Qe as the relation between T and Qs changes. If we fix (nu/V) as constant (i.e. maintain constant advance coefficient, J=V/nD and propeller geometry) the only way to increase force is to increase ne or Qe, both which increase engine horsepower.

[ 09-29-2002, 12:13 PM: Message edited by: John E Hardiman ]

Lowell Bernhardt
09-29-2002, 01:55 PM
WOW John! :eek: I took 24 credit hours of electrical engineering in college, and you just blew by me! I didn't think about it when I posted that last night, that, DUH you can't have one of any of those without the other.

Now leme see maybe I can try this in hillbilly termonology:

What ya wera tryin to say is, ya need a BIG OLE motor that could oughta run really loud to make um, the boat go fast?

(High HP=high torque) + high RPM = FAST

Right equation? Of course the variable in this equation would be the proper prop pitch and so forth.

mmd
09-29-2002, 02:24 PM
Eloquently stated, John. It always concerns me when the majority of boat owners and especially wanna-be designers naively state belief that powering a vessel is a simple, straightforward exercise, when in reality it is fraught with such expensive subtleties.

Lowell Bernhardt
09-29-2002, 07:44 PM
I wasn't trying to undermine John's approach. I was just wondering which boat would win under the set criteria. Just a simple question, a boat designer I'm not.

brad9798
09-30-2002, 08:24 AM
A 45 "fast" boat with TWIN 1,100 hp engines and only 60mph?

Look at something else ... that's not as fast as it 'should' be with 2,200 hp!!

I think of the Fountains and Bajas around me, and they go faster than that with less hp.

A 47' Fountain by me will run closer to 80mph with that hp. Of course, I believe its beam is only 8'6" or so. Plus, it's got that stepped hull design ... who knows.

Brad

Dan McCosh
09-30-2002, 08:49 AM
Theory aside, a high-torque, low-rpm motor tends to be pretty heavy, which is why you don't see many speed records set by tugboats. A lightweight, high-rpm engine is pretty much the only way to make a light, planing hull work, which is what you want for top speed.

Ian McColgin
09-30-2002, 09:47 AM
You'll be fine unless the horse power exceeds penile diameter (in millemeters) times IQ.

John E Hardiman
09-30-2002, 10:19 AM
Originally posted by Ian McColgin:
You'll be fine unless the horse power exceeds penile diameter (in millemeters) times IQ.Ian, in my experience hp is inversely proportional to IQ and directly proiportional to beer consumed so the dimensionally correct equation would be:

max hp =penile diameter (in millemeters)*beer consumed (in grams)* g/(754000000*IQ)

where g = acceleration due to local gravity in meters/second

:D

PilotArt
10-02-2002, 08:12 PM
A simple equation to convert torque to horsepower:

Ft.Lb. torque times RPM divide by 5252 equals horspower.

(1477x1600)/5252=450. :cool:

[ 10-05-2002, 09:50 AM: Message edited by: PilotArt ]