General design approach
an example
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General design approachan example |
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General
approach – an example Do you have perspiration running down
your forhead after reading about the general design approach? I had when I wrote
it! So lets take an example different
to Trinitus: I want a glider with a span of 2,8 meters (110”), my guestimated minimum effective flying speed is 7m/s and I choose the well proven airfoil SD 7037 which has a critical renoylds number around 80.000. My
wing chord at 90% will then be: Re-number
= V x C-90% x 70.000 - if
I move the numbers around a bit, I get: C-90%
= 80.000 / (6,5 m/s x 70.000) = 0,163 meters equals 166mm (about
6,5”). To
get a strong glider a want a thick wing at the root, so I choose a 260mm
for my root chord (C-root) Remember
tau?? This is where he gets into the story: (Ln
(C-tip) – Ln (C-root) ) / (-0,83) (Ln
(163) – Ln (260)) / (-0,83) = 0,561 All
we have to do now, is calculate the chords at different places along the
span: Remember the
formula: C-unknown = (cosine (arc-sine (unknown)))tau x
C-root In our specific example it converts to: C-unknown = (cosine (arc-sine (unknown)))0,561 x
260 We want a wing that is split in three – a flat center and tips with
tiplets. Balsa trees in Denmark do ... nomally NOT grow at all, but in
most shops they normally grow in 1 meter lengths, so
my center section will be 1000mm. This means that I want to know the chord 500mm
from the fuselage (half of 1000mm), since the tiplets are 10% of the
half span, they start 1260mm from the fuselage (1400 x 0,9), I know have 4 places
where I’m interested in calculating the chord. These places have to be
compared to the half span: Actual distance from the fuse:
Compared to the half span: 0mm
0 / 1400
= 0,000 500mm
500 / 1400
= 0,357 1250mm
1250 / 1400
= 0,893 1380mm
1385 / 1400
= 0,986 (this leaves me 15mm for the final tip shape) 1400mm
1400 / 1400
= 1,000 Insert these recalculated span wise numbers in formula number 3 , and
you get: Distance from the fuse: chord:
My wing: 0mm/root
260mm (255mm theese two chords can be changed to a rectangular constant chord middle section with a chord of 255mm (about 10"= 500mm
250mm 1250mm
166mm
165mm 1385mm
88mm
90mm 1400mm
0mm
30mm We
do now have a wing for a glider!! Calculate
the area, aspect ratio (4a), weight (5) and wing loading (6): Span:
2,8 meters Area:
60,7 sqdm AR:
formula 4a
12,9 Weight:
formula 5
1629 grams Wing
loading: ( 1629g / 60,7 sqdm = ) formula 6
26,8 g/sqdm Airfoil:
SD 7037 Now
is the time to have a look at the whole thing – the weight is
reasonable, meaning that it can be build strong enough at 1,6 kilo or a
little above, but how does the wing loading compare to the airfoil
camber?? Take a look at the table and notice that with this wing loading
the camber shall be around 3,0% and that’s exactly what the camber of
SD 7037 is. All we have to do now, is to determine if we
want ailerons, and/or brakes or flaps, and how much dihedral we need –
and last but not least -
put a fuselage on it and get out and catch some thermals. This
particular plan form is by the way the basis for an all wood glider Í
call ”Vivid L”, controls on rudder/elevator and flaps or brakes –
look at the three plan sketch or start designing your own super glider.
And take an extra look at the sketch, and let your memory wander back in
times – remember the ”Legend” from Airtronics? Legend and Vivid L
have almost identical plan forms, though mine is 10mm more narrow at the
2 tip chords. I didn’t cheat; I just reconstructed it. |
