How do I determine lift needed?

[orangexception]

Typically when I build planes, I just throw on wings and see what happens.

Is there a formula I can use or general guidance to determine how much lift I need for a plane?

I'm loosely thinking there's some kind of ratio between weight and lift.

Edited November 18, 2013 by orangexception
marking as answered

[Starwhip]

Typically when I build planes, I just throw on wings and see what happens.

Eyup. That's what I do too.

I'm loosely thinking there's some kind of ratio between weight and lift.

And yes, there is. I don't know it off the top of my head, but it is similar to TtWR (Thrust to Weight Ratio). I suggest the wiki or google.

[Smidge204]

I think the "lift" parameter of each wing part is in Kilonewtons, similar to "Power" (thrust) of rocket engines.

So a pair of Delta Wings, each with a lift of 1.9, would presumably generate a lift of 3.8kN. Dividing that by 9.81 gives us a total of 387.36 tons of mass that these wings could conceivably support.

Providing, of course, I'm right about the kN thing. I've already complained about that.

=Smidge=

[BurntPhoenix]

Since KSP doesn't give you the wing area, it's hard to determine how much weight you have per square meter of wing and therefore how well your aircraft will fly with the lift figures and thrust to weight ratio. If I were you, I wouldn't bother with the math - using the "guess and test" method is probably far more convenient and less time consuming.

Edited November 18, 2013 by BurntPhoenix

[fatfluffycat]

Why don't you just install MechJeb or kerbal engineer (Is that its name?) to tell you?

Don't mess with this "formula" and "math" nonsense

[Black_Tomcat]

I think the "lift" parameter of each wing part is in Kilonewtons, similar to "Power" (thrust) of rocket engines.

So a pair of Delta Wings, each with a lift of 1.9, would presumably generate a lift of 3.8kN. Dividing that by 9.81 gives us a total of 387.36 tons of mass that these wings could conceivably support.

Providing, of course, I'm right about the kN thing. I've already complained about that.

=Smidge=

I just have one small problem about your hypothesis:

Lift is relying on speed. So the formula to calculate it has to have V somewhere in there.

[Smidge204]

I just have one small problem about your hypothesis:

Lift is relying on speed. So the formula to calculate it has to have V somewhere in there.

This is absolutely true! Okay, then I have no idea whatsoever what the "Lift" property of the wing parts means.

=Smidge=

[numerobis]

ChevronTango tracked this one down.

[Thourion]

Yeah well, i poked at that thread, got dizzy, and im back to "Make it look sexy" mode.

@fatfluffycat: Jeb or KE, dont show lift, am i right? or did i miss something again?

[orangexception]

ChevronTango tracked this one down.

Oh ah, that's a nice read.

[orangexception]

Yeah well, i poked at that thread, got dizzy, and im back to "Make it look sexy" mode.

I also approve of this method.

[numerobis]

My take on this question is "trial and error works pretty well".

If you do the math out fully, you find that going 1 m/s down the runway, with the wings tilted 10 degrees up using shift-WASD in the SPH, every unit of lift provides about 0.14 kN of lift.

Multiply that by your speed in m/s: at 50 m/s you're getting 0.14 * 50 = 7 kN per unit lift.

Divide that by the air pressure (divide by e^{altitude/5000}, or, roughly, divide by 3 every 5000m): at 300 m/s at 10km in level flight you're getting 0.14 * 300 / e^2 = 5.74 kN per unit lift.

If you have a 20 degree angle of attack, the same basic number is 0.21 kN per unit lift, and at ~25 degrees it's at a maximum of about 0.22 kN. More than that and lift starts to fall, which means that you probably want to incline the wings 10 degrees and stay within 15 degrees of your prograde marker, or you gradually stall.

Edited November 18, 2013 by numerobis

[el_coyoto]

I took note of results I found in some thread (that I can't find any more) on my old trusty notebook dealing with the lift rating/mass ratio and its relation with rotate speed at take-off.

The main objective of the thread was to help people with building planes that take off before the end of the runway.

And although orangexception wasn't asking about take-off speed specifically, I still find it interesting to share, the logic being that a plane that has a reasonable take off speed *should* have enough lift and fly nicely.

The thread defined "lift rating" as the sum of lift number of aero parts divided by the mass of the plane.

The author provided numerical results which I reproduced fairly accurately on a few tests :

[table=width: 500, class: outer_border]

[tr]

[td]Lift rating[/td]

[td]Test result[/td]

[/tr]

[tr]

[td]0.36[/td]

[td]failed to take off, end of runway, 110 m/s[/td]

[/tr]

[tr]

[td]0.68[/td]

[td]Rotate @65 m/s[/td]

[/tr]

[tr]

[td]1.06[/td]

[td]Rotate @45 m/s[/td]

[/tr]

[/table]

So, if at some point in the SPH I wonder if my plane has a chance of taking off and flying without leaving the SPH, I quickly sum the lift of the biggest aero parts (laziness) my design uses and check if it has a lift rating of around 0.6.