Scale Models - View Single Post - SE5a

LINNEY · 22-01-2008

I have now studied the article, discovered and sent to me by Greyhead for which many thanks, and it is very interesting. The author of the article was seeking information to help make his model electric and he quoted data comparing electric power sources for a 71" span Hawker Fury bi-plane weighing 9.5lbs - sensible because it is almost identical the to author's own model. The author also said that he knew that his aircraft would fly well with a 4-stroke 90 glo engine - so presumably would the Fury.

I then went back to various "expert" opinions. One says that the cubic inch capacity of an i/c engine multiplied by 1000 equals the watts required of an equivalent electric motor so a 90 engine can be replaced by a 900 watt motor - apparently. But another "expert" says that an electric motor should yield 100 watts of power per lb of weight so the Fury needs 950 watts. Very similar. But another "expert" says you only need that much power for aerobatics. For calm steady and impressively controlled demonstration flying you only need 80 watts per lb, i.e. 760 watts in this instance.

So what about the Fury comparisons? Well, the author quoted comparisons with 12 different motors but different props too! Why different props? Baffled me, the aircraft needs a certain diameter prop to "propel" it down the runway and with a pitch and prop speed (hence mph speed) sufficient to give it lift-off - if you want to actually fly.

Anyway, the 12 different comparisons were interesting. I will not quote them all, just the extremes. In all instances they were run at a (theoretical) airspeed of about 50 mph, calculated from pitch and engine rpm. For anyone not usually interested, speed in mph is very roughly prop pitch in inches multiplied by rpm divided by 1000, hence an 8" pitch at turning at 7,000 rpm would give an airspeed of 8x7 = 56mph. More accurately such quick figures are about 5% high, not really significant since all this is theoretical, actual flying air speed will be affected by "prop slippage", a function of drag and prop design.

So, to the author's figures. Well, one motor turned a 24x12 prop at
about 4,200rpm with a 2:1 gearing. Thus the engine was turning at about 8,400 rpmand apparently using 45 amps at 36 volts, some 1,600 watts! Bit high by "expert" suggestions. But at the other end another useful motor turned a 17x8 prop at about 6,200rpm with a 3.7:1 gearing and hence a motor speed of about 23,000rpm! And power? Apparently 36 volts consumed 38 amps, hence about 1,350 watts. Still higher than expert predictions - but so what? (no pun intended) Am I seriously being told that if I need a 24" prop to drag an aircraft down a runway I can alternatively do it with a 17" prop if I rotate it 50% faster? By that reasoning put 12" diameter props on a Lancaster bomber and run them at a few million rpm and the aircraft will move down the runway. Really?

And another Fury test showed a direct drive Outrunner motor (which I intend) turning a 20x11 prop at about 4,700rpm used 36volts at 40 amps = about 1,400 watts. Sounds interesting, but the Fury weighs only 9.5lbs against my (anticipated) 15lbs and unfortunately has no quoted figures for wing loading - well, none that I can find. And currently I am replacinga 60 4-stroke on my 71" 8lbs Cub with a 13x8 prop at 7,000rpm with a direct drive outrunner using 800 watts! Very close to the middle of the "expert" figures (by chance!). I wish I could read the current consumption at minimum actual flying speed but such kit seems a bit too expensive for me.
This brings me back to my constant point. An aircraft, model or otherwise, will need a certain size of prop to drag it along, depending upon its effective weight, and a certain airspeed to enable it to lift off, this latter determined by wing-loading and drag.

I repeat (boring, no doubt) if someone can tell me what size prop is needed by a "draggy" biplane weighing 15lbs and what airspeed is needed to enable lift-off for the same aircraft with possibly a 24oz/sqft wing loading then I can work out pitch/rpm and find a power supply to match.

Failing that I shall await with eager anticipation Greyhead's actual data when he starts flying his i/c SE5a. THAT is the real way to find out.

22-01-2008	#23 (permalink)
LINNEY Scale Model Member Join Date: Jun 2007 Visit LINNEY's Gallery Posts: 20	I have now studied the article, discovered and sent to me by Greyhead for which many thanks, and it is very interesting. The author of the article was seeking information to help make his model electric and he quoted data comparing electric power sources for a 71" span Hawker Fury bi-plane weighing 9.5lbs - sensible because it is almost identical the to author's own model. The author also said that he knew that his aircraft would fly well with a 4-stroke 90 glo engine - so presumably would the Fury. I then went back to various "expert" opinions. One says that the cubic inch capacity of an i/c engine multiplied by 1000 equals the watts required of an equivalent electric motor so a 90 engine can be replaced by a 900 watt motor - apparently. But another "expert" says that an electric motor should yield 100 watts of power per lb of weight so the Fury needs 950 watts. Very similar. But another "expert" says you only need that much power for aerobatics. For calm steady and impressively controlled demonstration flying you only need 80 watts per lb, i.e. 760 watts in this instance. So what about the Fury comparisons? Well, the author quoted comparisons with 12 different motors but different props too! Why different props? Baffled me, the aircraft needs a certain diameter prop to "propel" it down the runway and with a pitch and prop speed (hence mph speed) sufficient to give it lift-off - if you want to actually fly. Anyway, the 12 different comparisons were interesting. I will not quote them all, just the extremes. In all instances they were run at a (theoretical) airspeed of about 50 mph, calculated from pitch and engine rpm. For anyone not usually interested, speed in mph is very roughly prop pitch in inches multiplied by rpm divided by 1000, hence an 8" pitch at turning at 7,000 rpm would give an airspeed of 8x7 = 56mph. More accurately such quick figures are about 5% high, not really significant since all this is theoretical, actual flying air speed will be affected by "prop slippage", a function of drag and prop design. So, to the author's figures. Well, one motor turned a 24x12 prop at about 4,200rpm with a 2:1 gearing. Thus the engine was turning at about 8,400 rpmand apparently using 45 amps at 36 volts, some 1,600 watts! Bit high by "expert" suggestions. But at the other end another useful motor turned a 17x8 prop at about 6,200rpm with a 3.7:1 gearing and hence a motor speed of about 23,000rpm! And power? Apparently 36 volts consumed 38 amps, hence about 1,350 watts. Still higher than expert predictions - but so what? (no pun intended) Am I seriously being told that if I need a 24" prop to drag an aircraft down a runway I can alternatively do it with a 17" prop if I rotate it 50% faster? By that reasoning put 12" diameter props on a Lancaster bomber and run them at a few million rpm and the aircraft will move down the runway. Really? And another Fury test showed a direct drive Outrunner motor (which I intend) turning a 20x11 prop at about 4,700rpm used 36volts at 40 amps = about 1,400 watts. Sounds interesting, but the Fury weighs only 9.5lbs against my (anticipated) 15lbs and unfortunately has no quoted figures for wing loading - well, none that I can find. And currently I am replacinga 60 4-stroke on my 71" 8lbs Cub with a 13x8 prop at 7,000rpm with a direct drive outrunner using 800 watts! Very close to the middle of the "expert" figures (by chance!). I wish I could read the current consumption at minimum actual flying speed but such kit seems a bit too expensive for me. This brings me back to my constant point. An aircraft, model or otherwise, will need a certain size of prop to drag it along, depending upon its effective weight, and a certain airspeed to enable it to lift off, this latter determined by wing-loading and drag. I repeat (boring, no doubt) if someone can tell me what size prop is needed by a "draggy" biplane weighing 15lbs and what airspeed is needed to enable lift-off for the same aircraft with possibly a 24oz/sqft wing loading then I can work out pitch/rpm and find a power supply to match. Failing that I shall await with eager anticipation Greyhead's actual data when he starts flying his i/c SE5a. THAT is the real way to find out.