View Full Version : Thrust vs. Horsepower
I don't know if this is the forum for this but... Does anyone out there know how to compare the power of a given electric motor to a regular outboard. The electrics all are rated in pounds of thrust. Whereas the regular outboards are in horsepower.
I am trying to decide on a new motor for river use (mostly up river against a gentle current).
09-05-2002, 09:40 AM
I have read that 1 horsepower equals about 22 lbs. of sustained thrust.
Thus, 5 hp = 110 lbs. of thrust
There are manufacturers (Minkot??) that make electric outboards, in the 200lbs of thrust range ... roughly equating to just over 9 horsepower.
09-05-2002, 10:18 AM
Interesting question. (Well, for an Engineer:) Found this website with numbers of 1HP ~= 65lb thrust. West Marine (http://www.westmarine.com/webapp/commerce/command/ExecMacro/news.d2w/report?article=nws_power42.inc)
09-05-2002, 10:20 AM
Minnkota has a FAQ that talks about HP and thrust:
Might be useful - or just confuse more <smile>
John E Hardiman
09-05-2002, 01:08 PM
The West Marine site is just winging in the dark about the effect of hull wake on thrust and the Minn Kota site is at least honest but deals only with effective hp (ehp), not shaft hp (shp) and never say the size of the "boat" used in their calculation.
So, setting super-cavitating and surface piercing propellers aside for now; the amount of thrust delivered by a propeller at a given horsepower is dependent upon the blade shape (mostly aspect ratio and sweep), the number of blades, and the advance coefficient J(J=Va/Dn where D=the propeller diameter, n=rps,Va= velocity of advance in fps). For a given J there is a coefficient of thrust (Kt) and a coefficient of torque (Kq) for any given propeller design. Kt and Kq are maximum at J=0 for most propellers in normal operating conditions. Kt goes to 0 (or negative) at J ~ Va/Pn[ :( edited to correct jeh 9/05/02] (P = geometric pitch). Kq remains positive at higher J than Kt but eventually goes negative and the propeller becomes driven rather than a driver (this is how wind generators work). The efficiency of a propeller is the ehp/shp and there is a single J where it is maximum. Ehp is calculated by the thrust(T)*Va/550. Shp is calculated by torque (Q)*n/550. Total thrust is calculated by T=rho*n^2*D^4*Kt and likewise total torque Q= rho*n^2*D^5*Kq. Note that to go twice as fast requires n to double to maintain optimum J, this leads to T and Q increasing by a factor of 4 giving an overall increase in the ehp and shp by a factor of 8!. Also note that by definition efficiency is zero at zero speed, but this is where the highest thrusts are developed. Of course, Va is not necessarily the speed of the boat or ship. If the propeller is in the wake of the hull, a significant amount of thrust can be gained because the Va at the prop is less than the Va of the hull. This is how VLCCs get by with relatively small power plants.
Now setting theory aside; you can expect the following thrusts at BOLLARD (Va=0)conditions for marine non-cavitating propellers:
Modern "series" propellers 30-35 lbs/shp 10+ hp
Modern "thrusters" 35-40 lbs/shp 5-2500 hp
Low Hp high aspect ratio trolling motors 50-60+ lbs/shp 1-5 hp
Very high aspect ratio human powered propellers 120-200 lbs/shp 0.75-1.25 hp
Modern speed boat propellers are generally of the cavitating nature and generally tend not to follow the above rules. Expect 50-100% of the thrusts for the modern series propellers above, but much higher J values.
[ 09-05-2002, 02:55 PM: Message edited by: John E Hardiman ]
09-05-2002, 01:24 PM
I printed your answer and filed it in my good stuff to know file.
Have you read a book called "To Engineer is Human"?
09-05-2002, 02:41 PM
From an electrical perspective...
Minnkota claims that my 74 lb. bollard thrust motor draws 46 Amps at full throttle and 24 volts. That's 1104 Watts or 1.48 electrical horsepower. Assume 85% efficiency (probably high) for the permanent magnet motor and you have 1.26 shaft horsepower. 58 lbs thrust per shaft horsepower is pretty close to John's estimate above. Isn't engineering fun?
By all means, read "To Engineer is Human".
John E Hardiman
09-05-2002, 03:07 PM
Note the edit to my previous post to correct an error as to the J when Kt goes negative. The offending offender has been castignated. smile.gif
Also note that if the hull form is very inefficient (i.e. high form drag), propeller efficiency can be greater than 100% when using Va based upon ship speed.
I hev to get better fingers
[ 09-05-2002, 03:08 PM: Message edited by: John E Hardiman ]
Thanks for all the replies. John I've printed your response and I'm going to endeavor mightily to understand it. :D Us accountants are good with plus's and minuses's but this higher math takes me back a few years.
I'll let you know what happens when I settle on a motor.
John E Hardiman
09-06-2002, 01:03 AM
Sorry I lost sight of what you were looking for. Here is what you need.
I assume you want to go slow with a displacement boat. Therefor we look at maximum hull speed without getting into too much wave making. This means that velocity through the fluid in knots (V) divided by the square root of length on the waterline in feet (L) should be between .8 and 1.0. This sets our design speed.
Next we need to determine our thrust required. Well, thrust should equal drag at design speed. Therefore if we knew our drag at the speed we wanted to go it would all be simple, however your hull may not have reams of tank test data. An approximation is required:
Thrust in pounds (T) =2* [0.00871+0.053/(8.8+L)]*wetted surface (S) in ft^2*V^1.825.
Now S can be approximated as .7* L*[(2*draft of canoe hull in feet)+ beam at the waterline in feet]
As stated before ehp =Va*T/550 where Va= velocity in fps (or Va=V*1.689) and for most applications shp is no greater than ehp/.6 and delivered hp (dhp) is approximately equal to shp/.8
L=25’; beamWL=6’; draft=2’: therefor S= 175 ft^2
Hull speed = 4-5 knots
Thrust required =67.8 lbs
Ehp = 67.8*5*1.689/550 = 1.04
Shp =1.04/.6= 1.73
Dhp= 1.73/.8= 2.16
Say a 3 hp motor
Of course I will deny any and all of this
John E Hardiman
09-06-2002, 02:06 AM
Reading back over my first post I notced a "sort of mistake" that could lead to some confusion. "n" is normally defined as revolutions per second (rps) in most american texts. The n given the equation for shp in my first post is for radians per second (also rps or more commonly omega in american texts). The ratio between the two is omega = 2*pi*n where n is in rotations per second. So:
Shp = Q*omega/550 = Q*2*pi*n/550
The equations for J, Kt and Kq are correct as long as you are consistant with the units; omega or n.
[ 09-06-2002, 02:08 AM: Message edited by: John E Hardiman ]
Georgian Bay Boy
09-10-2002, 12:19 AM
I have a MinnKota 12v.,55 lb. thrust trolling motor w/long shaft which I occassionally use on my c.1963 Lightning. In terms of electrical horsepower (ever seen an electrical horse?) I think of it as 0.9 hp. With crew and gear, total hull weight is over 1000 lbs. Hull length is 19', beam is 6'6" I think, draft is 5" and CB depth almost 5'. The motor is fine if its not fighting too much in the way of wind and waves. For threatening situations, I'd want a motor 2 or 3 times more powerful.
John, thanks for the info. Now I just have to finish designing the boat so I can make the calculations. The draft will be minimal (my goal is five inches or less) because the boat is intended for low river use. I know this is too shallow for most motors. Thats why I am interested in electric. The motor is light and I can lift it quickly when a rock or sandbar comes in sight. Hop out of the boat. Pull it past the obstacle and hop back in/on and lower the motor. Sounds like work and it is, but it will be easier with the new boat than the heavy aluminum I use now. The new boat will probably be a pontoon style with two 16' closed kayak like pontoons. I can calculate the speed of the river when it is low and will factor in overcoming that when going up river. You are quite right that I am looking for something relatively slow. When the river is higher and faster we will stick to the big ugly aluminum boat with a 35 hp two stroke on it.
09-10-2002, 01:16 AM
Don't you just love it when something goes click.
Like working out CLR on a boat.. balance the underwater profile on a pin...calculate the areas etc.. OR stand beside the boat at a dock and push it away with your foot until you find the middle.LOL.
Johns comment above "thrust equals drag". CLICK.
Bearing in mind a no wave hull speed.
You can build the boat, load it up and tow it with some scales in the line to get your thrust.A bit like you do with a dinghy to find the best towing point at a given speed. Or to compare 2 dinghies.
(I might be slow but I get there in the end)
John E Hardiman
09-10-2002, 11:39 AM
Originally posted by whb:
.. The draft will be minimal (my goal is five inches or less) because the boat is intended for low river use... <snip> ... The new boat will probably be a pontoon style with two 16' closed kayak like pontoons....Just a note about the thrust equation:
T =2* [0.00871+0.053/(8.8+L)]* S * V^1.825.
If the beam (b) divided by the draft (t) is much different than ~2-3, the factor of 2 at the beginning of the equation will change. So say ~3 for b/t=5 and ~1.5 for b/t=0.5.
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