Wow what a great answer in detail. I will go over it thoroughly the next few days. This is the kind of stuff I want to learn.
Thanks! Good luck in flight and life!
James Alderson
To: Indoor_Construction_at_yahoogroups.com
From: john.barker783_at_ntlworld.com
Date: Mon, 22 Apr 2013 21:50:01 +0100
Subject: Re: [Indoor_Construction] Re: F1M Design
James I am
not sure if you are just being casual when you mention multiplying mm by grams
or if you were looking for recommendations on calculating the CG. I apologize now if I am overegging
an answer because you are fundamentally correct , one adds up the moments and
divided by the total weights. I
give an example below.
I will assume
some typical figures for an F1M, (cg (lower case) refers to a component
cg))
Motor stick,
15.5” long, 0.65g, cg half way along, Front of motor stick is taken as the
Datum
Tail Boom,
17” long, 0.37g, cg half way along. (cg is 24” from the Datum)
Wing, 8”
chord, 0.81g, cg at mid chord, LE is 2” behind the Datum. (cg is 6” from the
datum)
Tail, 5”
chord, 0.40g, cg at mid chord. (cg is 30” from the Datum)
Propeller,
0.70g, ( cg assumed 1” in front of
the Datum),
The total
moments are: ((7.75x0.65)+(24x0.37)+(6x 0.81)+(30x0.40)+(-1x0.70)) = 30.0775
inch gram.
The total
weights are: (0.65+0.37+0.81+0.40+0.70) = 2.93g.
The CG is
(30.0775/2.93) = 10.265 inches behind the Datum.
This is just
behind the trailing edge of this theoretical model but I have not yet added in
the moment from the rubber motor (or the 0.07g ballast to bring the weight up to
3g!). You could, of course, run
through all the calcs again but including the motor weight of 1.5g as allowed by
the rules but it is usually more convenient to start with the airframe weight
and CG that has just been calculated and add in the rubber. The rubber will have its cg at half the
motor stick length behind the Datum so the new total moment is: (10.265”x2.93g)
+ (7.75”x1.5g) = 41.7 inch gram and the new Total weight is:
(2.93+1.5)=4.43g. So the CG with
rubber is: (41.7/4.43) = 9.413” behind the Datum. At about 93% of the wing chord. Just for interest; if the 0.07g ballast
needed to bring the model up to weight is added on the extreme end of the tail
boom them the CG will move to 101% of the wing chord.
Obviously
your weights and dimensions will be somewhat different but I hope this is of
some use. It is not easy to
tabulate calculations neatly on mailing lists.
John Barker -
England
Received on Mon Apr 22 2013 - 17:46:25 CEST
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