Re: Watts per Pound?

Here's a good rule of thumb for electrics. 50 watts per pound will
> fly like a Cub , 75 watts per pound will do some aerobatics , 100
> watts per pound will hover. This will vary of course , depending on
> the aircraft , but , it WILL put you in the ballpark.
> Hope this helps.
and 30W per pound will fly like a GWS tigermoth :-)
Its not *quite* that simple tho. The wttas per pound does two things -
overcomes the drag, and any surplus gets to lifte the mass of teh
aircraft against gravity - a (in my day) O-level mechanics exercise in
calculation. So heres one for you all. How many watts does it take to
lift one pound at 1000 feet per minute in an ideal frictionless world?
The reason parkflyers get way with less than 50W/lb is that they fly SLOWLY.
First order approximations have drag and lift roughly relaed - i.e.
double the weight needs double teh lift which is double teh drag.
Similar maths from your O-level course will show you that the power
needed is proportional to drag times speed. So to keep a lb of plane in
the air at 30moph takes roughly twice as many watts as to keep one in
the air (assuiming it doesn't stall) at 15mph.
The original '50W/lb' rule was formulated for sport planes with stall
speeds in the upper teens and lo twenties, with overall
battery/ESC/motor/gearbox/prop efficiences around 50% or less.
However broadly your original tables are correct..but I'll amplify tehm
a little
30-40W/lb Slofly or parkfly, or other low wing loading scale plane (GWS
Tiger moths etc)
40-60W/lb Good basic trainer, or draggy WWI biplane. Also parkfl;y
warbirds like GWS.
60-80W/lb Warbirds or smooth aerobats, probably good enough to loop from
level. This is where picojets and the like tend to start operating.
Aklso entry level gutless ducted fans :-)
80-100W/lb. Fast models with almost unlimited vertiocal, and possibly
hover potential with efficient setups, but not fully 3D. minimum for
decnt ducted fan IMHO.
100W-130W/lb. Full 3D in the right plane with unlimited vertical. Ducted
fand now starts to look like a jet.
i have alos derived another formula, which isn't quite as useful as it
was with the advent of LiPoly, and that is that the weight of motor and
cells etc. and teh equivalent power delivered is such that without the
airframe,. a typical setup is capable of delivering 200W/lb. That means
that to get e.g. 100W/lb the airframe weight must be no more than equal
to the weight of the flightpack and motor.
For e.g. 66W/lb the airframe is twice the weight of the power train.
This is a useful rule of thumb for electric conversions.

> Ken Day
> >
>
>>Hi;
>>
>>A while ago, I looked up the "Watts per Pound" rules, trying to decide on a
>>motor/prop/cell count package for my plane, and I can't remember, are the
>>Watts counted as "Watts input" or "at the prop"? Can you help me out? >>
>>To put the question in a context, here's my application: I need 30oz. of
>>"motor stuff" to balance 40oz. of airframe (in the shape of a GeeBee) at the
>>proper center of gravity. I'd like rise-off-ground capability at least, so
>>by the guidelines, I think I'll need around 60W per pound, or 260W total -
>>and I'm assuming that's output.
>>
>>I'm thinking it's going to be an AXI 2820/12 with a 10x6 on twelve cells...
>>According to the manufacturer, that will give me 253W (close enough) at a
>>cost of 23A...
>>
>>Have any better ideas? I'd love to hear them. Especially if I can get
>>decent duration and power out of a geared, cheap can. Frankly, I'm >>guessing...
>>
>>Thanks. A bunch.
>>Dan.
>>
>>
>
Reply to
The Natural Philosopher
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Well its an interesting diversion anyway.
It used to be the input watts per pound, at the motor terminals, cos that's where its easiest to measure, unless you have a dyno :)
With modern power train efficiencies up into 70% plus, its not that much different.
Reply to
The Natural Philosopher
It's "Watts in." While you can get a general idea of the Watts at the propeller, calculations are far from accurate (e.g. Motocalc results), and are far from easy to obtain. Input Watts, on the other hand, are simple: Multiply the number of cells by the number of Amps to get a reasonably accurate Watts figure under load.
Nope, that's the input, and as I stated in another thread, I don't think 60W/lb is enough for a GeeBee. The Watts-per-pound rules make many assumptions, including that the airplane has adequate wing area and is not exceptionally draggy.
You're closer than you were in the last thread. If it were me, I'd be shooting for 100W/lb input on a brushless system. Motocalc is really horrible in predicting the performance of the AXI 28XX series motors for some reason, so I depend on actual test data from
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On 12 cells, you can go up to a 12x8 prop without pushing the motor too hard, and have more than enough poop to move a 4lb GeeBee.
Actually, I might have an idea for you. A power setup I call the "Gary Wright Power System" might be ideal for this plane. It starts with a 10-cell Sub C pack, a Kyosho Endoplasma brushed motor ($20), a Great Planes GD600 gearbox (
Reply to
Mathew Kirsch
| So heres one for you all. How many watts does it take to | lift one pound at 1000 feet per minute in an ideal frictionless world?
To lift a pound weight up 1000 feet requires 1356 joules. 1356 joules/minute = 22.60 watts.
Show my work? :)
% units 2083 units, 71 prefixes, 32 nonlinear units You have: 1000 feet pound-force You want: joules * 1355.8179 / 0.00073756215 You have: 1355.8179 joules/minute You want: watts * 22.596965 / 0.044253731
(Yes, this is simply a problem of doing unit conversions, once you realize that a foot-pound is a unit of energy.)
Unfortunately, this exercise isn't very relevant for a plane. For a car (or elevator) it's pretty good, as a car has something solid and large (the Earth) to push against, but a plane's propeller is moving air around, which is a quite inefficient method of producing thrust.
However, it's more efficient to accelerate a large volume of air by a small amount than to accelerate a small volume of air by a large amount, which is why gear boxes and larger props are usuaully a good thing.
Reply to
Doug McLaren
Hey, thanks everybody (you in particular, Mathew).
I like the sound of the geared setup you mentioned, but I think I'll go with the brushless configuration for now: The fewer moving parts, the better, and I'm really shooting for "light", although the horse is already out of that barn :^)
I think it's going to be the AXI 2820/10, starting with 10 cells (I'd really like ten cells - I have a variety of chargers that can do ten cells. A whole drawer-full of "variety") and I may try a few different props in the 11x7 neighborhood, until I get off the ground...
Thanks a bunch! Give me about a month to get this tied to my plane somehow, and I'll give you a flight report (or a crash post-mortem). Dan.
Reply to
D
Interestung tho, as we reckon it takes 80W/lb, which suggests the overall efficiency of everything is 25%. I reckon 90% on the airframe 70% on teh prop 90% on teh gearbox and 80% on the motor...Mmm. A lot of room for improvement then?
And helicopters. Don'y forget helicopters. Huge diamter, low RPM and fine pitch :)
Reply to
The Natural Philosopher
| > To lift a pound weight up 1000 feet requires 1356 joules. | > 1356 joules/minute = 22.60 watts. ... | Interestung tho, as we reckon it takes 80W/lb, which suggests the | overall efficiency of everything is 25%. I reckon 90% on the airframe | 70% on teh prop 90% on teh gearbox and 80% on the motor...Mmm. A lot of | room for improvement then?
1000 ft/second really isn't that fast -- only about 11 mph. So if your plane can climb at a 30 degree angle, it only needs to go at 22 mph.
If a more efficient `elevator' is your goal, the easiest way to improve that efficiency is probably to climb faster. Afer all, if you're merely hovering (or flying level), your efficiency is 0%. :)
If you're looking for more efficient thrust generation in an airplane in general, orinithopters (i.e. flapping wings) are probably the way to go.
| > However, it's more efficient to accelerate a large volume of air by a | > small amount than to accelerate a small volume of air by a large | > amount, which is why gear boxes and larger props are usuaully a good | > thing. | | And helicopters. Don'y forget helicopters. Huge diamter, low RPM and | fine pitch :)
Actually, the wing on a plane serves the same purpose. It's purpose is to take a lot of air and push it down gently, which pushes the plane up. A small wing has to push the air down faster to generate the same lift, and since the kinetic energy imparted to this air is porportilan to the velocity squared, it's less efficient.
This is why sailplanes have long wings -- so they can push a lot of air down slowly. The longer the wing, the lower the induced drag. Of course, the down side is that the large your wing, the higher the parasitic drag ...
Personally, I find this stuff fascinating. If I had it all to do again, I'd have studied aerospace engineering in school. Designing (or even better, flying) planes sounds better than mucking in front of a computer all day (though I'm sure they spend more than their share of their time in front of a computer nowadays too.)
Reply to
Doug McLaren
1000 ft/second is 11 mph?
Reply to
Randy A. Hefner
| > 1000 ft/second really isn't that fast -- only about 11 mph. So if | > your plane can climb at a 30 degree angle, it only needs to go at 22 | > mph. | | 1000 ft/second is 11 mph?
Sorry. 1000 ft/minute is what I meant, but I suspect you knew that.
(:
Reply to
Doug McLaren
I came up with 22.6 watts. Close?
Art
Reply to
Art K6KFH

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