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Boat Forum / Building / July 2006Electric outboards
I have noticed that real numbers concerning electric outboards seem to be few and far between. Therefore, I thought some of you might be interested in some results I obtained the other day with mine. I used two Excide 6 volt golf cart batteries in series. Open circuit voltage at the time of the tests was 12.53 volts. The motors were both Minnkota Enduras, one 30 lb, the other 50 lb. I have no way of directly measuring thrust, nor did I have a calibrated current shunt, so I assumed the factory-published values of 30 lb at 30 amps and 50 lb at 42 amps. Voltages at the input to the motor leads were 12.14 and 11.96 with the motors set to max. This indicates and combined internal battery and external wiring resistance of about .013 ohm. The nominal input powers are therefore .49 hp and .67 hp. Again, assuming factory stated thrust is accurate, the output powers are .28 hp and .53 hp for overall efficiencies of 57% for the 30 pounder and 80% for the 50. Incidentally, when I questioned Minnkota by phone they would not state efficiencies but did say the 50 lb unit is their most efficient. These motors push a 16 ft flat-bottomed sailing skiff at 3.5 and 4.0 mph respectively. I would be pleased to hear from any of you who are also interested in electric propulsion. Bob Swarts > I have noticed that real numbers concerning electric outboards seem to > be few and far between. Therefore, I thought some of you might be > interested in some results I obtained the other day with mine. > I used two Excide 6 volt golf cart batteries in series. Open circuit > voltage at the time of the tests was 12.53 volts. > The motors were both Minnkota Enduras, one 30 lb, the other 50 lb. I > have no way of directly measuring thrust, nor did I have a calibrated > current shunt, so I assumed the factory-published values of 30 lb at 30 > amps and 50 lb at 42 amps. > Voltages at the input to the motor leads were 12.14 and 11.96 with the > motors set to max. This indicates and combined internal battery and > external wiring resistance of about .013 ohm. The nominal input powers > are therefore .49 hp and .67 hp. > Again, assuming factory stated thrust is accurate, the output powers are > .28 hp and .53 hp for overall efficiencies of 57% for the 30 pounder and > 80% for the 50. Incidentally, when I questioned Minnkota by phone they > would not state efficiencies but did say the 50 lb unit is their most > efficient. > These motors push a 16 ft flat-bottomed sailing skiff at 3.5 and 4.0 mph > respectively. See the extra power needed for that 0.5 mile faster? > I would be pleased to hear from any of you who are also interested in > electric propulsion. The Queen Mary II has electric propulsion.;-) For now I think battery propulsed boats will be a niche product, nice for fishing or cruising protected areas or coming home but not for holidays. Do the math and find out yourself. My boat has also electric propulsion. When the day comes that fuelcell or other electric generation becomes cheaper then a diesel generator set, I'm prepared. Untill then the diesel keeps running.  SignatureRichard e-mail: vervang/replace invalid door/with NL.net http://web.inter.nl.net/users/schnecke/ >> I have noticed that real numbers concerning electric outboards seem to >> be few and far between. Therefore, I thought some of you might be [quoted text clipped - 35 lines] > cheaper then a diesel generator set, I'm prepared. Untill then the > diesel keeps running. Bear in mind that many of the boats on the water today ARE used for fishing or utilitarian functions on protected waters. No one suggested serious cruising. But my 16-footer does have a 30 mile range at 4 mph at the 70% discharge level which is more than adequate for its crabbing and homecoming (sail backup) functions. The math does work in many cases, but, as you say, picking the right niche is important. BS >I have noticed that real numbers concerning electric outboards seem to >be few and far between. Therefore, I thought some of you might be [quoted text clipped - 26 lines] > >Bob Swarts Small electric motors can have 80% efficiency. Small water propellers can have 80% efficiency. If you take the product of voltage across the motor, AT the motor, and the current through the motor, and multiply by 0.64 and divide by 746 you'll have an estimate of the net HP available for thrust. V x I x 0.64 / 746 = HP for thrust. Thrust at constant power varies with water speed, and is greatest at standstill (which is why troll motor makers specify thrust at standstill, where it is meaningless) Let's work your numbers: 12.14V x 30A x 0.64 / 746 = 0.3 HP 11.96V x 42A x 0.64 / 746 = 0.43 HP Rough, rough cross check: if power required is proportional to v^3 then power required at 3.5 mph is 3.5 x 3.5 x 3.5 / ( 4 x 4 x 4) or 0.67 of power at 4 mph. Power available for thrust at 3.5 mph = 0.3/0.43 = 0.88 of power at 4 mph This suggests to me (it could be a dozen other things) that the prop is less optimal on the faster skiff. Take it with a pinch Brian Whatcott Altus OK >> I have noticed that real numbers concerning electric outboards seem to >> be few and far between. Therefore, I thought some of you might be [quoted text clipped - 56 lines] > > Brian Whatcott Altus OK Basically I agree with your analysis, but you assume equal over all efficiencies for both motors, which negates both Minnkota's published specs and my observations. Modern small electric motors can (not saying this is actually the case)have efficiencies well above 90%. One of the problems is that neither Minnkota nor MotorGuide will state efficiencies for their motors or props. But IF this is the case with the Minnkota 50 lb model, then an over all efficiency of 80% would not be impossible. In any case, without being able to accurately measure the current or thrust at speed, it is difficult to wring much more information from my results. Also, there were probably a couple hundredths of a volt more drop along the internal and external motor cable (I measured at the input end of the factory cable). Note, though, that your theoretical values of available hp (.3 and .43) are not in bad agreement with my observed results of .28 and .53 hp. I sure would like to see some more people contribute some actual numbers. BS With a GPS and a spring scale, you could get a pretty good approximation of the thrust. Attach a drag; e.g. a triangle of plywood, to a line and measure both the reduction of speed and the pounds of drag. Assume as a first approximation that the thrust/drag curve is about linear for a small increment of drag. (We know it's exponential, but you're probably not up near hull speed where it goes vertical.) If there's any wind or current, make runs in several directions. Add drag in increments and you'll generate a series of slopes. Find an exponential curve that fits and you should have a pretty good guess at the thrust curve. Roger (or maybe not) derbyrm@NOSPAMinsightbbNOSPAM.com http://home.insightbb.com/~derbyrm >>> I have noticed that real numbers concerning electric outboards seem to >>> be few and far between. Therefore, I thought some of you might be [quoted text clipped - 75 lines] > > BS > With a GPS and a spring scale, you could get a pretty good approximation of > the thrust. Attach a drag; e.g. a triangle of plywood, to a line and [quoted text clipped - 91 lines] >> >> BS I am not sure this will work satisfactorily. The drag total drag force is comprised of that from the plate and that from the boat itself. I could measure the force on the plate as you suggest with a spring scale, but unless it is significantly greater than the drag on the boat, I can't solve for the thrust. And while it may be much greater than the drag on the boat as its size increases and the speed decreases, that will be at low speed, so I still don't have thrust at full speed. BS >> With a GPS and a spring scale, you could get a pretty good approximation of >> the thrust. Attach a drag; e.g. a triangle of plywood, to a line and [quoted text clipped - 13 lines] >> >> "Bob S" <ravensrest@olypen.com> wrote >I am not sure this will work satisfactorily. The drag total drag force >is comprised of that from the plate and that from the boat itself. I [quoted text clipped - 5 lines] > >BS Work with him, on this one. You are going [let us say] at 3.44 mph via a gps speed average of ten runs, and then you drag a plate which drags at 4.23 lb for an average of ten readings, which reduces your gps speed to 3.21 mph average of ten readings. Because drag and thrust are in balance (else you speed up or slow down) you can say it takes 4.23 lb of thrust to contribute the extra 0.23 mph from 3.21 mph Although this one set of data is not enough, in principle you could say T (thrust) proportional to 3.21 ^2 kT = 3.21 ^2 and also, T + 4.23 proportional to 3.44 ^2 k(T + 4.23) = 3.44 ^2 Supposing that the scaling constant k stays constant over this small range, we can divide one equation by t'other to get this: (T + 4.23) / T = 3.44 ^2 / 3.21 ^2 = 1.148 T + 4.23 = 1.148T 0.148T = 4.23 T = 28.6 lb. at 3.21 mph Like that, but with more data points. Brian Whatcott Altus OK >>> With a GPS and a spring scale, you could get a pretty good approximation of >>> the thrust. Attach a drag; e.g. a triangle of plywood, to a line and [quoted text clipped - 47 lines] > > Brian Whatcott Altus OK Thanks for your input, Brian. Unfortunately k is not constant. The problem is that T1 = [k-boat + k-plate]*v1^2 If we eliminate the plate T2 = T1 + del-T = [k-boat]*v2^2 Dividing, we get T1/[T1 + del-T] = {[k-boat + k-plate]/k-boat}*{v1/v2}^2 I can measure del-T as you suggest as I can the velocities, and solve for T1. But I can not accurately predict the individual drag coefficients or their ratio. One might assume k-plate is much, much less than k-boat (resulting in your equation), or assume it is some particular per cent of k-boat, but then I am not sure that the result would be any better than assuming the factory-rated thrust is realistic. And, of course, we also have to face up to the inaccuracies of the speed and thrust measurements. Conversely, one could use two different sized plates, each of which had much, much greater drag than the boat, assume the drag differs only by the ratio of the plate areas, and solve. But again, I am afraid this is no more accurate than the factory spec. To get the plate drag much greater than the boat drag, you would also be in a very low speed regime where the static thrust might be as good as whatever one measures. Static thrust could be measured fairly accurately with a spring scale by pivoting the motor on its mounts from a stationary point(a dock)and measuring the force needed to hold the motor vertical. BS You're getting away from what I suggested. My idea was that the drag of the plate was much much less than that of the boat. Measuring at various speeds got the slope of the curve, dV/dD at those speeds. (Velocity and Drag) It's a backward, graphic approach to integral calculus and you have the advantage of knowing the type of curve. Also, if your voltmeter is reasonably precise, you do have an ammeter. Just record the voltage drop across the cables feeding the motor. Again, you don't have nice text book values unless you put a known load on at some point, but you can get the percentage change in current. (The voltage drop between no-load and your load is proportional to current and to the relatively constant resistance of the battery, connections, and cables. Ohm's law you know.) Roger derbyrm@NOSPAMinsightbbNOSPAM.com http://home.insightbb.com/~derbyrm >>>> With a GPS and a spring scale, you could get a pretty good >>>> approximation of the thrust. Attach a drag; e.g. a triangle of [quoted text clipped - 80 lines] > > BS > You're getting away from what I suggested. My idea was that the drag of the > plate was much much less than that of the boat. Measuring at various speeds [quoted text clipped - 97 lines] >> >> BS I have to leave town for a couple of days, but I'd like to follow up on this. First, I think Brian and I both made a mistake in confusing thrust and drag. In our approaches we actually allowed the thrust to change by the amount of the drag on the plate. The correct assumption, it now seems to me, is that thrust remains approximately constant for small changes in drag (say, 5 lb of added plate drag when the total thrust is 50 lb). Then T = K*v1^2 and T = (K + dK)*v2^2 The result is dK/K = (V1/v2)^2 - 1 and we can determine the ratio of the drags, but not the thrust. This could be, however, a useful tool for determining hull drag if we know the drag on the plate. How accurately can one predict drag on a fixed size and shape plate? Getting back to your idea: I can see that you can make a series of pair-wise measurements over a small range of drags that give the approximate slope of the velocity-drag curve at each measurement point. Graphically connecting these straight line sections would approximate a small portion of the velocity-drag curve, but I do not see how one gets thrust from it without knowing the actual drag. BS If one is not accelerating (or decelerating), then thrust equals drag. Roger derbyrm@NOSPAMinsightbbNOSPAM.com http://home.insightbb.com/~derbyrm >> You're getting away from what I suggested. My idea was that the drag of >> the plate was much much less than that of the boat. Measuring at various [quoted text clipped - 127 lines] > > BS > If one is not accelerating (or decelerating), then thrust equals drag. > > Roger > derbyrm@NOSPAMinsightbbNOSPAM.com > http://home.insightbb.com/~derbyrm Yes, of course, but both Brian and I allowed thrust to increase by the amount of the added plate drag. That is not correct. BS >>> You're getting away from what I suggested. My idea was that the drag of >>> the plate was much much less than that of the boat. Measuring at various [quoted text clipped - 126 lines] >> >> BS Hmmm? (I'm to lazy to dig into the math.) With only small changes in drag and the throttle held constant, it would seem to me that the thrust of the motor would be constant. Since the thrust balances the total drag one gets the dV/dD which should be the same as dV/dT, (Velocity, Drag, Thrust) but I assumed that dT was zero. I do see the dilemma. Did somebody divide by zero here? Okay. If thrust is constant for the small change in drag, then total drag is constant, and we're getting the equation of speed versus hull drag; i.e. the reduction in speed correlates to reduced drag by the hull. That works. (I think.) Roger derbyrm@NOSPAMinsightbbNOSPAM.com http://home.insightbb.com/~derbyrm >> If one is not accelerating (or decelerating), then thrust equals drag. >> [quoted text clipped - 139 lines] >>> >>> BS Nice post! Just as practical as it could be. Brian Whatcott Altus OK >With a GPS and a spring scale, you could get a pretty good approximation of >the thrust. Attach a drag; e.g. a triangle of plywood, to a line and [quoted text clipped - 91 lines] >> >> BS > Small electric motors can have 80% efficiency. Small water propellers > can have 80% efficiency. > If you take the product of voltage across the motor, AT the motor, and > the current through the motor, and multiply by 0.64 and divide by 746 > you'll have an estimate of the net HP available for thrust. > V x I x 0.64 / 746 = HP for thrust. > Thrust at constant power varies with water speed, and is greatest at > standstill > (which is why troll motor makers specify thrust at standstill, where > it is meaningless) Thrust depends on the propeller dimension. Bollard pull (Google gives plenty hits) is commonly used with tugs, but conditions have to meet minimal requirements to measure. My previous boat was (minimal) faster with its 11 hp then a identical boat with 16 hp. Bollard pull with my boat give a 150 kg, the boat with 16 hp 100 kg. > Let's work your numbers: > 12.14V x 30A x 0.64 / 746 = 0.3 HP > 11.96V x 42A x 0.64 / 746 = 0.43 HP > Rough, rough cross check: > if power required is proportional to v^3 > then power required at 3.5 mph is 3.5 x 3.5 x 3.5 / ( 4 x 4 x 4) or > 0.67 of power at 4 mph. > Power available for thrust at 3.5 mph = 0.3/0.43 = 0.88 of power at 4 > mph Thrust and boat speed don't behave proportional, why else bollard pull for tugs. > This suggests to me (it could be a dozen other things) that the prop > is less optimal on the faster skiff. Possible indeed. I think efficiency is best tested by measuring the supplied power at different speeds, but requires an ammeter. SignatureRichard e-mail: vervang/replace invalid door/with NL.net http://web.inter.nl.net/users/schnecke/ |
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