Former Member Posted March 23, 2018 Share Posted March 23, 2018 [This posting has been removed] Quote Link to comment Share on other sites More sharing options...
Dickw Posted March 23, 2018 Share Posted March 23, 2018 Your mixing up linear measurements e.g. wingspan with square measurements e.g wing area. If the tailplane needs to be 20 - 22% of wing area it needs to be 155 to 171 sq.in in area as you say, but you can’t apply the 20 – 22 % to the span or chord. You need to work those out from the tailplane area. For a 3:1 aspect ratio stab you will need something like 22.5” span and 7.5” in chord (= 168.75 sq.in area). Good luck with the project, it sounds interesting. Dick Quote Link to comment Share on other sites More sharing options...
Former Member Posted March 23, 2018 Author Share Posted March 23, 2018 [This posting has been removed] Quote Link to comment Share on other sites More sharing options...
Denis Watkins Posted March 23, 2018 Share Posted March 23, 2018 Dick's beat me to it Don't lump the stabiliser and the elevator together The elevator is 20% - 30% of the stabiliser The stabiliser being 15% - 20% of the wingspan The fin area 33% of the stabiliser And the rudder 33% - 50% of the fin These are ballpark numbers If you have graph paper, or just paper ST, draw out your model to a scale to fit the paper and errors will show up visually Then you can look again until the drawing looks right Edited By Denis Watkins on 23/03/2018 16:28:23 Quote Link to comment Share on other sites More sharing options...
PatMc Posted March 23, 2018 Share Posted March 23, 2018 Tailplane + elevator [UK models don't have horizontal stabs] = Chord x span Span = chord x aspect ratio = 3 * chord therefore Area = chord * 3 * chord = 3 * chord² therefore Area/3 = chord² Required area = 155 Therefore Chord² = 155/3 = 52 [aprox] Chord = √52 = 7.25 [aprox] Span = 3* chord = 21.75 Ditto if area = 171 Chord = 7.6 Span = 22.8 Edited By PatMc on 23/03/2018 16:38:48 Quote Link to comment Share on other sites More sharing options...
Former Member Posted March 23, 2018 Author Share Posted March 23, 2018 [This posting has been removed] Quote Link to comment Share on other sites More sharing options...
The Wright Stuff Posted March 23, 2018 Share Posted March 23, 2018 Posted by supertigrefan on 23/03/2018 16:17:42: Ah!....I see So I need to start with the required area, then play with the span and chord until I get it at the 3:1 AR? Indeed, although with a tiny bit of algebra, there's no need to play. If the tail plane span is x, then the area is x * x/3 if the 3:1 ratio is to be respected. So Area = (x^2)/3 or (transposing): x = squareroot(3*Area) If the required area is 155 sq. in. then the span is squareroot(3*155) = 21.56 inches. If the required area is 171 sq. in. then the span is squareroot(3*171) = 22.65 inches. Hope that helps. TWS EDIT: Crossed posts, PatMc. Great minds, eh? Edited By The Wright Stuff on 23/03/2018 16:40:40 Quote Link to comment Share on other sites More sharing options...
Dickw Posted March 23, 2018 Share Posted March 23, 2018 Posted by Denis Watkins on 23/03/2018 16:23:20:..... The elevator is 20% - 30% of the stabiliser The stabiliser being 15% - 20% of the wingspan .............. Edited By Denis Watkins on 23/03/2018 16:28:23 Are you talking areas or linear measurements here? We don't want to further confuse the OP . Dick Quote Link to comment Share on other sites More sharing options...
Martin Harris - Moderator Posted March 23, 2018 Share Posted March 23, 2018 While we're clarifying things Dennis, you wrote "Don't lump the stabiliser and the elevator together". Did you mean that the elevator shouldn't be included in tailplane area calculations? I don't claim any more than casual designing experience but I've always considered the elevator to be part of the tailplane when calculating a C of G. Edited By Martin Harris on 23/03/2018 16:57:37 Quote Link to comment Share on other sites More sharing options...
Denis Watkins Posted March 23, 2018 Share Posted March 23, 2018 Deary deary me, well the numbers were taken straight from here **LINK** The diagrams are giffs that I cannot take across to here, but they make sense written with the aeroplane in view Edited By Denis Watkins on 23/03/2018 17:09:33 Quote Link to comment Share on other sites More sharing options...
Former Member Posted March 23, 2018 Author Share Posted March 23, 2018 [This posting has been removed] Quote Link to comment Share on other sites More sharing options...
PatMc Posted March 23, 2018 Share Posted March 23, 2018 Your original numbers don't add up. 66" span with a 5:1 AR gives 871sq ins. 66" span with an area of 776 gives an AR of 5.6:1 Perhaps best if you state the parallel section chord & parallel sub span + the tip chord. BTW 5:1 is quite a low AR e.g. the Piper Cub is 7:1 also 20% tailplane (inc elevator) area is pretty generous but the distance between tailplane & wing needs to be taken into account when deciding this %. Edited By PatMc on 23/03/2018 17:46:53 Quote Link to comment Share on other sites More sharing options...
Martin Harris - Moderator Posted March 23, 2018 Share Posted March 23, 2018 Posted by PatMc on 23/03/2018 17:39:15: BTW 5:1 is quite a low AR e.g. the Piper Cub is 7:1 also 20% tailplane (inc elevator) area is pretty generous but the distance between tailplane & wing needs to be taken into account when deciding this %. This is an important consideration to bear in mind when you are being quoted "ball park" figures. The effectiveness of a tailplane is a product of its area times the distance from the centre of pressure so the further from the wing the tailplane is, the smaller area is required. An extreme example is a tailless model where the allowable C of G will be very close to the leading edge of the wing - a very small percentage of the mean chord as opposed to the 25 - 35% ranges often quoted for "typical" models. Edited By Martin Harris on 23/03/2018 18:07:26 Quote Link to comment Share on other sites More sharing options...
Former Member Posted March 23, 2018 Author Share Posted March 23, 2018 [This posting has been removed] Quote Link to comment Share on other sites More sharing options...
Martin Harris - Moderator Posted March 23, 2018 Share Posted March 23, 2018 Close - but I make it 771.25 (287.5 + 483.75) sq. in. wing area. Edited By Martin Harris on 23/03/2018 18:39:49 Quote Link to comment Share on other sites More sharing options...
Former Member Posted March 23, 2018 Author Share Posted March 23, 2018 [This posting has been removed] Quote Link to comment Share on other sites More sharing options...
PatMc Posted March 23, 2018 Share Posted March 23, 2018 What Martin said & the AR = 5.65:1. Quick way to calculate AR when the span & area are known : Span² / area = AR. Quote Link to comment Share on other sites More sharing options...
Former Member Posted March 23, 2018 Author Share Posted March 23, 2018 [This posting has been removed] Quote Link to comment Share on other sites More sharing options...
PatMc Posted March 23, 2018 Share Posted March 23, 2018 Sounds fine. Quote Link to comment Share on other sites More sharing options...
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