quick PofF question

[spitfire747]

hello all, doing my FI course at goodwood and have been asked the following question

if the speed for the best lift to drag ratio in the cruise is say 85kts for the tommahawk, then why is the speed for the best lift to drag ratio in a glide 70kts and not 85kts...?

any ideas
Cheers
Spitty

[Jinkster]

Anything to do with a feathered prop - this would alter the L/D ratio

Good question spitty!

[Send Clowns]

That does not sound right. There is a slight difference (in a glide the lift is slightly less than weight, so speed will be lower) but to a first approximation the best Lift : Drag speed should be constant, as it depends only on angle of attack (about 4° for a normal aerofoil). Why do you want to know the best Lift : Drag in the cruise? Is that for endurance speed? Not flown the PA-38 but I would be surprised if it was that high.

[Mass Times Velocity]

If L/D ratio incorporates thrust then lack of thrust will reduce L/D ratio (one element of drag (profile) being overcome by element of thrust), open to correction,
MTV

[Send Clowns]

Not sure what you mean by that, M x V. In level flight thrust overcomes all parts of drag. In the glide thrust overcomes none of the drag. This does not alter the drag at a given speed and angle of attack. Lift to drag ratio doesn't incorporate thrust, just lift and drag, so lack of thrust will certainly not affect the lift to drag ratio at a constant angle of attack.

[High Wing Drifter]

Here is my guess.

In the cruise you need to maintain about 4deg AoA for best lift/drag qualities of the aerofoil, so the speed is set accordingly. In decent, as in the climb, lift is less than weight so the best glide/climb speed (Vy) will be slower...at Vimd for a prop (if I remember correctly).

[Alex Whittingham]

My guess would be that the data in the question is wrong.

this (http://www.eaa1000.av.org/technicl/perfspds/perfspds.htm#Maximum_Range) is all I could find on the net. It suggests best range is about 90kt and best lift:drag about 70kt.

[Send Clowns]

Doesn't fit, HWD. The reduction in lift in the descent is only drag x sin (angle of descent). Sin (angle of descent) is, for small angles, approximately tan (angle of descent) which is in fact the reciprocal of the glide ratio. For example for a PA-28 this would be about 1/11 of the drag. Since lift is in that case 11 times drag (as lift to drag ratio is also the glide ratio), you are only losing 1/121 of lift. I think. Never tried this calculation before, but it seems sound. Anyone find a flaw?

[High Wing Drifter]

Shows what happens if I try and do something from memory without having finished the course - hehe. I have to say you have confused me a little though :)

I had a quick look at the notes on my laptop. Best range is of course Vimd. Best climb for a prop will be somewhere between Vimp and Vimd as the power available from thrust falls rapidly as you climb. For glide, the speed is Vimd because that is the best EAS over Drag ratio and there is no thrust other than gravity in the equation.

Assuming your 11/1 glide ratio then glide ang = inverse tan(11/1) = 85 deg. So the glide angle is 5deg. 1000kg * cos(85)(or *sin(5)) = 87kg less lift.

A tad more than 1/121. Hell! I'm not sure if this is right or not. Anybody else?

:confused: := :O

[Mass Times Velocity]

Send Clowns,
that would have had to be my 13th post!! Didn't manage to articulate my thoughts very well but your further posts have certainly given me something to think about, head in the books for me for the weekend to get principle really straight,
MTV

[Send Clowns]

HWD

You are calculating the lift reduction from the lift, it is actually a component of the drag (as it is the upward component of the drag, which is upslope, that reduces the lift requirement). Shame on you for taking lift in kg :ooh: (my physics tutors would spin in their graves at the units JAA force us to use, if they were dead), but to simplify I will keep it the same. The drag would be about 1000 / 11 = 91 kg. The component of this upward would be 91 x sin (5 degrees). This is about 8 kg. To complicate the issue the forward slope of the lift vector means that roughly 4 kg extra lift is required, meaning a net reduction of the order 0.4% in the lift required.

Again I have to add "I think". This is well beyond the ATPL syllabus, drawing on half-remembered first-year mechanics. I may well have missed out another factor, as at these levels approximations start to show their flaws.

MTV

Don't worry - not only have I done the whole course but I taught it too, although that was some time ago. This is beyond what you will need to know!

I find it interesting though. Sad :(