RCM&E Special Issue 2012

[Peter Beeney]

I have to say to be absolutely Pedantic about it all, strictly speaking the term foot-pounds here is incorrect. It should be pounds-foot. Strange as it may seem, there is a difference.

Also I would like to consider it permissible to talk in newton centimetres, N-cm, if I may, as apposed to kilogram centimetres. If I’m feeling frivolous I say there are 10 newtons to the kilogram, but on the other hand if I’m really serious about being pedantic I say there are 9.8066500286 newtons to the kilogram. A bit masochistic, too, perhaps, I really will have to throw the Abacus out one day and get a proper calculator!

I tend to get confused by all this. What exactly are N-cm’s? I think it is a unit, or moment, of torque and as such does it really tell me a great deal? What I think I also really need to know is in what sort of time this operation is carried out in. Will it be in hours, minutes or seconds? Looking a one or two examples the standard seems to be the speed of operation over 60 degrees, one sixth of 360 degrees in fractions of a second. So can we now do some calculations of the power output of the servos?

So what do we know? We know the rotary power output of our servo is the torque multiplied by the revolutions. Very conveniently one N-cm per second is exactly equal to one watt, so we just need to quantify our units in equal terms. So we can use these figures to look at two fictional servos. Both have a quoted output of 50 N-cm torque, but looking closer one has a quoted time of 0.25 sec per 60 degrees, the other has a slower time of 0.5 sec per 60 degrees. So finding the power of the first example, a, we can say that that one revolution will take 6 times 0.25 sec = 1.5 seconds. This equates to 0.75 revs per second. Power = torque by revs per second, so 50 by 0.75 = 37.5 watts. Therefore the servo a has an output of 37.5 watts
Servo b says that one revolution will take 6 by 0.5 = 3 seconds. This equates to 0.333 revs per second. 50 by 0.333 = 16. 666 watts. Therefor servo b has an output of 16.666 watts.

Based on these figures servo a is more than twice as powerful as servo b, but you have to check the small print to find that out. On their own do the torque figure mean very much? I’m not sure, but like the electric motor the power output can only every be the torque multiplied by the rpm. Like the motor when it is stalled for whatever reason, if there are no revolutions there is no output power, all the watts are expended as heat in the motor. This makes sense because the servo is driven by an electric motor anyway.
However, in normal circumstances, using these figures for time of transit it’s easily possible that it would be concluded that servo a was indeed considerable more powerful than b without doing any calculations.

I too have often wondered about the Harvard. A friend built the Flair version a long time ago, with a Saito four stroke. He flew it once or twice but had some difficulty, particularly landing it. So he gave it to me. It was by no mean a heavy model, I’m sure I would have done the wing loading, I can’t remember now, but it definitely would have been very light. But even so, even on only a medium paced landing it tended to drop a wing without warning, which could catch you unawares.
So I’ve often wondered if the larger ones were the same, and now it seem as though they might possibly have been to some extent. But I guess that with modern transmitters and two aileron servos it’s an easy matter to dial in a few millimetres of reflex, up aileron, mixed with low throttle. Just enough to improve the washout at low speeds.

PB

Edited By Peter Beeney on 28/09/2012 14:39:51

[Foamie Dave]

Have I missed something?

[Biggles' Elder Brother - Moderator]

Posted by Peter Beeney on 28/09/2012 14:27:44:

Very conveniently one N-cm per second is exactly equal to one watt, so we just need to quantify our units in equal terms.

No it not. Your analysis is based on the "work-done times the time it was done in" equals the power. Which in itself is sound. But the torque isn't the work done.

Torque and work done happen to have the same dimensions (units) but they are not interchanagable because they are fundementally different physical things. A standard "trick question" given to students applying to do engineering at university is "Torque has units of Newton.metres, work done has units of Newton.metres. So why are they different - why have two names for the same thing?"

The correct answer of course is that they are not same thing. Work done (force times the distance the force moves through) is a scalar quantity - ie it has magnitude but not direction. Torque (force times the leverarm distance it acts over) is a vector quality - it has magnitude and direction - and is only completely defined by both elements. The two are completely different things - they just happen to have the same units. You can't take the torque and say its equal to the work done and then multiply iy by the time and say that its the power - it doesn't physically make sense!

If you did want to do that then it would be torque times the angular distance over which the torque was active - that would be the rotatry work done!

BEB

[Ian Jones]

Well...

...anyway, I've been wanting some guidance on setting up a bungy bungee launch system for some time and there it is all in one nice neat article. The DIY foot release is so simple, I wish I'd thought of it and the advice for safe operation is invaluable.

Well done Dave Royds.

Edited By Ian Jones on 28/09/2012 16:04:28

Edited By Ian Jones on 28/09/2012 16:05:18

[Biggles' Elder Brother - Moderator]

Hear hear! I might get that EDF jet now!

BEB

[Foamie Dave]

Thanks for the kind words chaps.

I used to hate the idea of a bungee until I had a try...now Im hooked ..erm literally

[David P Williams]

WRT the Harvard's flying characteristics - I used to work on full size aircraft, mostly vintage types - Austers, Tiger Moths, Stearmans etc. including one Harvard. I was often the 'observer' on CofA or CAA permit test flights, one section of which was stalling in various configurations - clean; take-off flap; landing flap and undercarriage. The Chief Engineer/test pilot was a VERY experienced, unflappable chap (ex-Barnstormer etc, etc) but even he struggled with the vicious wing drop at the stall that almost rolled it on its back.

David

[Peter Beeney]

Just checking BEB, I did wonder if someone would spot my deliberate mistake. What I actually should have said is that 1 newton centimetre per second is equal to 0.01 watt, so that now makes the rotary output of servo a = 0.375W and that of servo b = 0.16666W.

Sorry if I misled you, I must learn to get the decimal point in the right place…..

David - Thanks for the Harvard observation, I’ve often thought that maybe scale model tend to slightly emulate their bigger brothers, and here that might be so. My friend had a Cub which he used to putter with, he thought the Harvard would be same. It wasn’t!

Now we need to know all about the Stearman Kadet….. please.

PB`

[David Ashby - Moderator]

Posted by Peter Beeney on 28/09/2012 17:56:05:

 

David - Thanks for the Harvard observation, I’ve often thought that maybe scale model tend to slightly emulate their bigger brothers, and here that might be so. My friend had a Cub which he used to putter with, he thought the Harvard would be same. It wasn’t!

 

 

What? We're talking about the full size Harvard Pete, the model flies very well indeed, albeit not for the beginner.

 

 

Edited By David Ashby - RCME on 28/09/2012 18:10:49

[Garbo]

Can I be a tiny bit critical? Harvard free plan, 30 photos and not a single one of its construction, equipment fit or engine instillation.

[Peter Beeney]

David, I’m sure the model does fly perfectly. All I’m saying, and as in the above post, too, I had one, a Flair, that to all intents and purposes should have been a doddle, and provided it was pressing on a bit, it was. But if it slowed down, normally, not excessively slow, on the odd occasion one wing would drop very suddenly indeed, generally the left as I remember, and I was never able to really say why this was; and of course I was always very conscious that I could be near this state. It’s now beginning to sound as though the big one also has some odd habits.

Maybe this was a one off, and all other model Harvard’s are fine, I’ve no idea. But, as I say, I’m sure that Tony’s is absolutely splendid. I’m certainly not inferring that it is not, I’m sure it’s more than well tried and tested.

PB

[Biggles' Elder Brother - Moderator]

Posted by Peter Beeney on 28/09/2012 17:56:05:

Just checking BEB, I did wonder if someone would spot my deliberate mistake. What I actually should have said is that 1 newton centimetre per second is equal to 0.01 watt, so that now makes the rotary output of servo a = 0.375W and that of servo b = 0.16666W.

Sorry Peter - we're still not there!

You see, power is not torque times revs/sec - it is ,in SI units;

Power (in Watts) = Torque (in Nm) times Angular Velocity (in radians per sec)

Dealing first with the torque - as you correctly amend - 50Ncm = 0.5Nm

Now turning to the angular velocity; 60 degrees is 60x2Pi/360 radians

Which is 1.05 radians.

So, for servo A: Angular velocity = 1.05/0.25 rads/sec

= 4.2 rads/sec

While for servo B Angular velocity = 1.05/0.5 rads/sec

= 2.1 rads/sec

So the mechanical power of Servo A is;

Power(A) = 0.5x4.2 = 2.1W

While for Servo B the mechanical power is:

Power(B) = 0.5x2.1 = 1.05W

An interesting extention to this is to note the following: suppose our servo is 60% efficient then the electric power which would need to be supplied to, say servo A, to deliver this mechanical power would be;

Power(E) = Power(M)/0.6

For servo A's 2.1W that is an electrical power of 3.5W. Now assumming that our servo supply is 6V then this power will equate to a current of;

Current = Power(E)/Volyage

or Current = 3.5/6 = 600mA

So, surprisingly, even at full deflection, at maximum capable torque the servo only draws a relatively modest current given the typical size of our poser packs. Interesting.

BEB

[John Olsen 1]

Poser packs?? Would that be what the Aussies call budgie smugglers?

I'll get my coat....

[Ian Jones]

Posted by Biggles' Elder Brother - Moderator on 28/09/2012 16:13:58:

Hear hear! I might get that EDF jet now!

BEB

'eck we'll have you on a slope next!

[Ponty Bri]

The special Autumn issue is packed full of great articles and technical info. I could particularly relate to Lee Smalley's article 'Getting back' more issues like this full of technical stuff and tips would be most welcome.

Brian

Edited By Biggles' Elder Brother - Moderator on 29/09/2012 21:09:45

[Biggles' Elder Brother - Moderator]

Posted by Ian Jones on 29/09/2012 16:02:08:

Posted by Biggles' Elder Brother - Moderator on 28/09/2012 16:13:58:

Hear hear! I might get that EDF jet now!

BEB

'eck we'll have you on a slope next!

That will never happen Ian. Last night I was looking at John Privett's video of his trip to a local slope. It was full of all these people wrapped up like Michellin Men with face masks, Tx gloves, the lot. I've seen Antartic explorers wearing less protection! They were struggling to stand upright because of the wind. And I just thought - this is a voluntary activity?!!! If we did that to prisoners we'd be up before the European court of human rights!

BEB

[Bernard Dolinski]

When will it be on the shelves on australia?

[Lima Hotel Foxtrot]

Posted by Garbo on 28/09/2012 18:44:11:

Can I be a tiny bit critical? Harvard free plan, 30 photos and not a single one of its construction, equipment fit or engine instillation.

Hear hear! In my honest opinion, while lots of pics of the completed model is fine, surely one or two could be swapped out for build/installation pics?

[Pete B]

Re above, I agree more build pics in the mag, please, but it's worth pointing out that there are more Harvard build pics than you can shake a stick downloadable as a PDF from this page

Pete

[slippyr4]

Surely shahid should know that servo torque is very much not measured in "kilograms per cm" !!!!

Edited By David Ashby - RCME on 01/10/2012 14:39:35

[slippyr4]

Posted by Peter Beeney on 28/09/2012 14:27:44:

If I’m feeling frivolous I say there are 10 newtons to the kilogram, but on the other hand if I’m really serious about being pedantic I say there are 9.8066500286 newtons to the kilogram.

I'll see your pedantry, and raise you that you're still wrong - A newton is never equivalent to a kilogram, because a kilogram is a unit of mass. What you mean is that at standard gravity, a 1 KG mass is accelerated by 9.8066 m/s/s, and hence has a weight of 9.806 N.

<bait>
Since our models fly at some altitude, they experience reduced gravity, and therefore our servos are less torquey in the air*
</bait>

[Peter Beeney]

Thanks for answer to the servo, BEB, I think my system of guessing failed yet again. Looks as though I’m going to have to take up knitting for a while. I’ll make something and then decide what is when I’ve finished. And that could be yet another guess…… A bit like my model aeroplanes!

I reckon I’m going to shut the old laptop lid for a time, or perhaps even somewhat longer than that. I think I need to get out more. I’ve been sitting around too long, got stuck in a rut, as they say.

PB

[Lee Smalley]

Guys can we keep this thread on track please, as riviting as this lesson is on units of measurement, save it for another thread where all you anoraks can argue together in peace

[Lee Smalley]

on the build pics issue, it is quite difficult to get the photographer to come round mid build and take some piccys, the alternative is to take it to the patch mid build and take them there, again we risk damage in transport, we generally can not take them ourselves as our cameras very rarely produce photos of acceptable quality! so it can be more trouble than you would think getting photos mid build

[Martyn Johnston]

Lee, you keep talking about aeroplanes while we're trying to discuss units here .

I've just got as far in the magazine as Shahid Banglawala's article and slippry4 is right; Sahid not only writes the units down wrong ; he actually states that "servo torque is measured in kilograms per centimetre (kg/cm)" .