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 Boat Design Forums HP to WATT to THRUST

#1
02-17-2005, 09:37 AM
 yipster designer Join Date: Oct 2002 Rep: 1114 Posts: 3,458 Location: netherlands
HP to WATT to THRUST

at the p-subs forum i found this:
Quote:
 It takes 746 watts to develop one horsepower. Watts are Volts multiplied by Amps. Volts times Amps equals Watts Divide by 746 equals Horsepower For example: 12 V X 40 A = 480W / 746 = .64 HP This is in a perfect world. You may want to down rate the actual HP a bit to account for electrical loss in the motor. Probably about 10%. Here's a link that may help you. http://www.wisegeek.com/what-is-horsepower.htm
its someting i was ewondering bout, now how does that compare again to "thrust"?
#2
02-17-2005, 09:54 AM
 Raggi_Thor Nav.arch/Designer/Builder Join Date: Jan 2004 Rep: 711 Posts: 2,457 Location: Trondheim, NORWAY
Electric outboards seem to produce more thrust per watt than gas outboards. I think it's because they have more propeller area per watt,.
__________________
Regards, Kvedja, mvh,
Ragnar Thor Mikkelsen
www.MBOATS.no
#3
02-17-2005, 11:03 AM
 jehardiman Senior Member Join Date: Aug 2004 Rep: 1909 Posts: 1,753 Location: Port Orchard, Washington, USA
Thrust is a function of the propeller, not the driving machinery.

The propeller absorbs hp as torque. Torque * rpm / appropriate constant = shaft hp (SHP). Now hp is absorbed by the boat as drag. Drag * speed / appropriate constant = effective hp (EPH). Now to make the boat go at a given speed, drag must equal thrust, so you can see that SPH and EPH are linked by the ratio of input torque to output thrust at a given speed and rpm for a specific propeller. These relationships are standard Kt (thrust coefficient), Kq (torque coefficient), and J (advance coefficient Va/(nD) n being rps). The propeller efficency (eta) is a measure of the ability of the wheel to change SHP to EHP eta= (J/2pi)*(Kt/Kq).

That said, it must be remembered that hp is absorbed by the prop as torque. Some driving machinery is better suited to delivering torque at the required rpm than others, i.e. direct coupled diesels and phase locked primary motors vice high speed gasoline IC or gas turbines. Better overall efficency can be had by despensing with any reduction gears.

Last edited by jehardiman : 02-17-2005 at 11:05 AM. Reason: spelling....Bad engineer! no cookie!
#4
02-23-2005, 02:15 PM
 Ssor Senior Member Join Date: Jan 2005 Rep: 10 Posts: 174 Location: Bel Air, Md
Horse power is a measure of work equal to 33000 foot pounds per minute. To determine the actual horsepower needed to move a boat at a given speed, towing is probably the most accurate means. If you know the speed in feet per minute and the force on the towing line then determining to hp is easy. The difference between the determined required hp and the engine hp reflects the efficiency of the transmission, shaft bearings, and propeller.
#5
02-23-2005, 02:41 PM
 cyclops Senior Member Join Date: Feb 2005 Rep: 38 Posts: 1,059 Location: usa
Electric motors are at least 2 X more eff. than IC engines. At 10 % and 90 % points of operation they can be 5 to 10 X more eff. You can not change the cam in the IC. You can instantly change the cam- " volts and amps. " in a electric.
#6
02-23-2005, 07:54 PM
 CDBarry Senior Member Join Date: Nov 2002 Rep: 295 Posts: 526 Location: Maryland
You have to do the propeller calcs using Jt, Kt, Kq curves or whatever.
#7
02-23-2005, 08:37 PM
 yipster designer Join Date: Oct 2002 Rep: 1114 Posts: 3,458 Location: netherlands
later alright? and i asume in dave Gerrs propeller handbook i want for my birthday. "Thrust is a function of the propeller" yes think i got that and the rest, what if trust is expressed as kg or lb? is that used only in air? is that comparable and to calculate back to Hp?
#8
02-24-2005, 09:24 AM
 Ssor Senior Member Join Date: Jan 2005 Rep: 10 Posts: 174 Location: Bel Air, Md
The small electic trolling motors are rated, Pounds thrust. Consider that oars are the ultimate low speed, large blade area boat propulsion systems. Further consider that a strong man can produce about one fifth of one horsepower, it follows then that high velosity thrust, i.e. jetskis, is not efficient for moving low speed boats. Paddling a canoe is a low effort, low power consumption means of moving a large load. But motorizing a canoe always involves an engine many times more powerful than a man. Reason: loss of efficiency.
#9
02-24-2005, 10:40 AM
 yipster designer Join Date: Oct 2002 Rep: 1114 Posts: 3,458 Location: netherlands

in a new quik search i saw a prop for a florida everglade flatbottom runner that was mentioned to be as good as doing 4 lb to the hp. so i asume thrust is a variable that needs CDBarrys reply "You have to do the propeller calcs using Jt, Kt, Kq curves or whatever."
#10
02-25-2005, 06:28 AM
 CDBarry Senior Member Join Date: Nov 2002 Rep: 295 Posts: 526 Location: Maryland
No, Gerr's book doesn't use a J, Kt, Kq approach. You use J (the ratio of speed to diameter and RPM) and the prop parameters to get the thrust coeficient, Kt and torque coefficient, Kq, which vary with J. Then you get the thrust and torque from RPM, diameter, and whichever K.

You can find these factors and the formuals to use them in professional texts like PNA. Larsson and Eliasson have some as well.

SNAME sells a program and manual called PSOP that produces them as well, and you can get the basic papers that they were originally published in through SNAME as well.

However the "take home" facts are that thrust and torque both vary with boat speed and RPM, falling to 0 when the boat goes fast enough, and at a maximum at zero speed but that the prop will only spin as fast as the available torque allows.

Also, the prop is limited by cavitation.

So you have to do the calcs and also say at what boat speed you want to know the thrust.
#11
02-25-2005, 08:46 AM
 yipster designer Join Date: Oct 2002 Rep: 1114 Posts: 3,458 Location: netherlands
thanks again CDBarry, took me a little comprehending.
from the pulse-jet forum i got this -different- formula from Irwin E. Treager's book ( i'll resize later today ) for a jet engine.
not in Gerr's book and as you mention other factors have to be included, sure interesting, thrust aint that easy to calculate i see...
#12
02-26-2005, 06:07 AM
 FAST FRED Senior Member Join Date: Oct 2002 Rep: 836 Posts: 3,606 Location: Conn in summers , Ortona FL in winter , with big dock & room for O'nite stop .
To simplify a "rule of thumb" is most propellers will produce 20 lbs of push for each HP.

Get the prop really optimized and maybe 25lb of push (THRUST) at certain speeds.

Every 10hp will cost a galon of gasolene per hour , the diesel will give 16 hp to 24 per gallon.

At 16 hp per gal your chugging an old DD ( 6-71 or similar) , at 24 hp per galon you have the latest electronic tripple shot injection ,waste gate dual turbos and inter or aftercooler high tech.

Which one would you prefer to be on AFTER the lightning strike?

FAST FRED
#13
02-26-2005, 08:07 AM
 Ssor Senior Member Join Date: Jan 2005 Rep: 10 Posts: 174 Location: Bel Air, Md
Fred, your "rule of thumb" for fuel comsumtion is pretty close but your thrust per hp doesn't take speed into account. We are stuck with hp=550 ft lb /sec,so 20 lb thrust would have to move the boat 550/20=27.5 feet in one second. so that 10 hp would move the boat 275 feet per second times 60 seconds in an minute =16500 feet divided by 6020 feet per nautical mile= 2.74 miles per minute or 164 kts. But if by reducing the speed we increase the thrust, we can have 200 pounds of thrust at 16.4 kts or 800 pounds of thrust at 4.1 kts. But to achieve this we must increase propeller diameter and torque.

Last edited by Ssor : 02-26-2005 at 08:24 AM. Reason: error
#14
06-01-2005, 02:07 PM
 John David Registered Join Date: Jul 2004 Rep: 13 Posts: 29 Location: Long Island
If one knows the prop shaft HP vs. shaft rpm* and the prop pitch,the theoretical thrust vs boat speed should be easy to calculate. Unfortunately this number must be reduced by propulsion efficiency. I have not been able to find any solid numbers for this, even from the prop mfrs. Suppliers of ball bearings for outboard engines list such efficiencies as 60 to 65%.This is then used to determine the load on bearings. Sounds
low to me but it is the only number I've got, so far. Does any one out there know what a typical outboard prop eff. is? In addition, does this number include loss due to lower unit drag (less skeg?) It seems to me it is not possible to measure such eff. without some mechanical equiv. of a lower unit. Hope to get an answer, John David

*This is derived from the engine Hp vs engine shaft rpm through the lower unit gear ratio.
#15
06-02-2005, 06:29 AM
 CDBarry Senior Member Join Date: Nov 2002 Rep: 295 Posts: 526 Location: Maryland
The prop mfgrs don't give efficiency because it varies with boat speed to prop RPM ratio (basically advance ratio, J). However, most props are standard enough that they can be predicted by one of the standard fits, usually Blount-Hubble. These use a regression on J, P/D, EAR, and Z (number of blades).
But it is always important to remember that the torque the prop absorbs can be no more than the engine puts out, so this is a big part of the matching issue.