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Old 01-26-2009, 10:31 AM
  #27  
Yak 52
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Join Date: Jan 2009
Location: UK
Posts: 64
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This post’s a long one hope you can bear with me!

This is what I came up with:

Full Size Aircraft................... Model

Scale 1/8.5556

Span - 25’ 8” or ..308” ............36.00”
Length - 22’ 1”... 265” ............30.97”
Height - 9’ 1.5” .....109.5” ..........12.80”
Chord - 4’ 6” .......54” ................6.31”
Gap - 4’10” ...........58” .................6.78”
Stagger - 2’ 3 ” .....27”................. 3.16”
Dihedral - 4.5 deg
Incidence - 2 deg
Port wing wash in - 2.25 deg

Tail span - 8’ 4.5” ...100.5” ...............11.75”
Airscrew diameter - 8’ 6.5” ..102.5” ...11.98”
Wheel track - 5’........ 60” ...................7.01”
Tyres - 700 x 75mm ...........................81.8 x 8.8mm
.......or 27.56 x 2.95”.......................... 3.22 x 0.35”


Areas:................................Model Area
Scalar: 73.19753

Wing area (S) - 212.1 ft2 ....2.89764 ft2 ...(417.25998 sq in)
Ailerons (each) - 11.6 ft2 .....0.15848 ft2 ...(22.82112 sq in)
............(total) - 46.4 ft2 ......0.63390 ft2 ...(91.28160 sq in)

Tail Area - 13.4 ft2 ...............0.18307 ft2....(26.36208 sq in)
Elevators - 12.2ft2 ................0.16667 ft2... (24.00048 sq in)
Total Tail - 25.6 ft2 ............0.34974 ft2... (50.36256 sq in)
Fin - ............2.2 ft2 ..............0.03006 ft2 .....(4.32864 sq in)
Rudder ....... 6.3 ft2 ..............0.08607 ft2...... (12.39408 sq in)
Total Fin - ..8.5 ft2 .............0.11612 ft2..... (16.72128 sq in)


Weights:
Scalar: 626.2553

Weight (W) - 1010lb......... (1.6127608 lb).... 29.0296944 oz


A few comments… (If I’m teaching granny to suck eggs just ignore me!)

I find it doesn’t pay to get too bogged down in the figures - it’s best to try and keep a practical understanding of what the figures stand for and constantly check on yourself to avoid mistakes in the calculations.


In terms of choosing what level of accuracy to work to (i.e. how many decimal places) what is the practical approach? Well, for example, I can design the chord to be accurate to 8 decimal places (6.31168831 inches) but can I cut my ribs to one hundred millionth of an inch? You might be able to if your kit is going to be laser cut!

So… I can cut to 1/100th of an inch or 0.25mm (sometimes!) so 2 decimal places are fine for linear measures. If I’m multiplying (e.g. areas) I’ll add a couple of decimal places to increase my accuracy so as not to end up with a too much compounded error.

Thinking of it the other way around - at this scale, moving something 1/100th of an inch on your model represents moving it less than 1/12th of an inch on the full size!

It does depend on the measure you are using though - 1/100 of a square foot is 1.44 square inches, so maybe a greater degree of accuracy is called for.

I actually found that the scale of 1/8.5556 gave me closer to the 36” span you were after.


The other practical thing to do is to keep checking against the drawings. For example, in your figures ‘Gap’ came out a smidge larger than Chord. Check the drawings - Yep that looks fine!

Stagger came out as pretty much half the chord. Check the drawings - perfect!


However… (… there’s always a however!)


Something seems to have gone a bit wrong with your wing area?

When you think about it, if span is 36” then a 34 sq in area indicates a chord or less than half an inch for wing… something drastically wrong there! (It may well be a typo!)

Again, there’s a practical check on that - multiply Chord by Span (x 2 - it’s a biplane!) to give you a ball park figure to aim at…

So working in feet (very roughly!):
3 x 0.5 (6”) x 2 = 3 ft2
From the specs: 2.89764 ft2
Just about right! (It’s less because the check didn’t exclude cut away part of the rounded wing tips)

If you want it in inches then multiply by 144 = 417.26 sq in

So the practical understanding will show you when something is wrong - it’s up to you to work out what that is though!


What’s the wing loading then?

Wingloading (W/S):
Wing area (S): 2.89764 ft2
Effective area (85%) 2.46299 ft
Target AUW (W) 29oz
So............ 29 / 2.463 =

Effective W/S = 11.8 oz/ft2

This is pretty encouraging - about what a conventional trainer/sport model would be I should think. You could probably go as high as 15+ oz/ft2 so that gives you a fair bit of head room with your target weight, i.e. up to your estimate of 38oz.

Andy Lennon’s article on Biplanes is worth a read:
http://findarticles.com/p/articles/mi_qa3819/is_/ai_n8799373

Part II has some comments on Biplane wing loadings:
http://findarticles.com/p/articles/mi_qa3819/is_199807/ai_n8802339



Dihedral at 4.5 degrees seems quite a lot but I would guess that is because rudder area is so small (there’s a direct relationship.)

The rudder/fin areas do seem pretty small (as is usual on WWI fighters) You might want to increase the size of the tail surfaces slightly…
It might be worth checking tail volume coefficient on that one. You’ll have plenty of aileron though! Pretty much a quarter of the wing area.

This site has some excellent FAQ info on that side of things:
http://www.djaerotech.com/dj_askjd/

I’d have to check but I think the reason for having the 0.25 degrees wash in on the port wing is designed to increase lift on that wing in order to compensate for the torque effects of the radial engine on the full size.


Hope some of that helps…

I’ve now spent so much time looking at the DH5 I’m seriously considering building one myself!

Jon

Last edited by Yak 52; 01-26-2009 at 10:43 AM. Reason: format
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