DD (not 34DD)
Your Q3
“ explain why we use a rudder to pick up a wing at the incipient stall without mentioning the use of ailerons (harder) 15 marks”
Just a thought, but the “without mentioning the use of ailerons” could be key
IF that is what the question said then I read it quite differently to the others who have posted on the topic. I see it as an attempt to get you to write about dihedral effect and explain why the use of rudder will make an aeroplane roll.
(Could be just me though. Glad I don’t have to pass the exams these days)
Q1 IMHO You have been given the points for this (less weight and less drags various) The reason that these ADVANTAGES are not used by many aircraft is because there are disadvantages but they don’t ask for those so lets not get into that.
Q 2. When you sideslip the wing will be a tad less efficient (reduced thickness chord ratio) So to generate the same lift as before you will have to use a bit more angle of attack – assuming you do not increase the speed – IE yank the nose up.
Q 4 There is but one effect of sweepback on induced drag (ID) – it increases it.
As there are 10 marks for this, they probably expect you to punt an explanation.
ID is of course totally lift dependent. No lift no ID. Lift being the upward force on the wing resulting from less pressure above it compared to that below it, THIS PRESSURE DIFFERENTIAL HAS TO ACT AT 90 DEGREES TO THE WING SURFACE. (apologies I’m not shouting just don’t know how to do the clever bold stuff etc) Because of this the more you increase the angle of attack the more you tilt the lift force backwards and increase its drag element.
Now to add the sweepback bit
For a given wing aerofoil section, the greater the sweepback the thinner the thickness chord ratio becomes (that is why we use sweepback of course – to kid the air it is seeing a thinner wing and so delay the onset of Mach effects without actually making the wing thinner structurally or reducing valuable internal volume) All other things being equal reducing the thickness chord of a wing will reduce the slope of the Lift versus AOA curve. A reduced slope in turn means that you need more AOA for any desired increase of lift. I.E. a swept wing needs more AOA for a given lift than a straight one. So the lift vector gets tilted backwards more on a swept wing hence more induced drag
Slasher.
Agree the sense of your words but I would not be happy to go into bat with your formula for ID!
From my words above you will see that I would be looking for an expression that calculated the rearwards (along the flight path) component of lift. For a symmetrical wing aerofoil this would be Lift x Sine AOA plus a term for the tip losses. For a more normal non symmetrical section you would have to add a little to the AOA value to take account of the actual zero lift angle
Of course I find I am wrong more and more often these days.
JF