Will my spaceplane be able to take off? How to figure.

[miganders]

First off, I'd like to give you a fair warning. Some math is involved. Thankfully, it's not that complicated.

For the sake of simplicity some factors have been left out. Most importantly drag and passive lift. This guide mainly applies to fairly aerodynamic crafts.

First, figure out your planes mass, thrust and steerable lift at takeoff.

Steerable lift meaning the lift of wings you are able to flap. Passive lift from static wings also matter, albeit less so than steerable wings.

We'll call this m, t and l.

Next, do the following calculations:

(m/l)*t = a

(t*l)/m = b

a/b = c

If c is in the range of approximately 1 to 35, your plane has too much lift and will spin out of control.

If c is in the range of 35 to 55, your plane should be able to take off fine.

If c is more than 55, your plane is too heavy and you'll have problems taking off before hitting the end of the runway.

Note that all of this is based on limited in-game testing, if you have results that are strongly contradictory, please let me know and I'll try to figure out why and incorporate it into this guide.

Edited September 1, 2012 by miganders
Added info about static wings.

[miganders]

There's a bit of a problem with this it seems.

Mainly that I am a terrible mathematician and the thrust doesn't actually matter in my current equation.

[vexx32]

Well, simplifying the equation gives (m^2)/(l^2). That's it, really. If trying to incorporate thrust too... well, technically any equation with the three interacting can be used as a guide if the interaction between the three is clear, and you have some experimentally determined guides of values that are good or bad.

I'm thinking a relationship like (l*t)/m would be sufficient to show the relationship between the three. Zero thrust means zero takeoff. Logical. Zero lift also means zero takeoff. Also logical (well, almost; if we define it as controllable lift, it can takeoff with zero controllable lift, but it's not easy to do, and it will not be able to be controlled very well if at all). The heavier you get, the more lift and thrust you need to take off. I think this one is logical as far as I can see. I don't know any useful numbers for it, but I can get some for you. To a point.

Problem is, this does not take into account where the controllable lift is (nose, tail, middle, etc.) or its orientation (e.g., a rudder wouldn't really contribute much to the takeoff part of things). But still, a useful guide. Well, let's see... One of my favourite planes that is capable of taking off before half the runway is gone, and then going vertical and flying upside down immediately has these values:

Thrust: 150.0

Mass: 6.99

Controllable Lift: 2.8 (I've ignored the control surfaces on the rudder here.)

Plug those numbers into (l*t)/m gives...

2.8*150/6.99 == 60.09

With this formula, higher number gives a better chance of takeoff. This thing can takeoff with practically no trouble, so I'd say a 60 is well into the realm of takeoff being ridiculously easy. However... not sure how it'd go on something with a ton of engines... Might wanna give that a try, if you would. Let me know if this works or not.

[EndlessWaves]

if we define it as controllable lift, it can takeoff with zero controllable lift, but it's not easy to do, and it will not be able to be controlled very well if at all

Actually it's perfectly possible to have an easily controllable machine with no control surfaces, it's just a case of building a stable airframe and using the vectoring jet engines as steering. It's obviously not as quick to respond as control surface equipped aircraft but it can be equally tame.

The following doesn't handle well as it's an extreme example but here's the machine I built to demonstrate lifting over 100 tons on a single 100 thrust jet engine:

It springs off the ground at just 17m/s and has zero control surfaces. It will keep flying straight and level until the fuel runs out (several hundred km later I'd imagine, although I've not tested it).

Controllable lift probably isn't the best name as canards produce a turning force rather than variable lift. On the runway this can have the same effect - with the back wheels braced against the runway a turning force will lift the nose up rather than rotate about the centre of mass.

Thrust determines the speed you need to take off. We have speed = thrust/weight.

But that's horizontal speed and only holds as long as your engine is horizontal, for lift off speed the formula is something like:

(horizontal thrust component / weight) * (lift + vertical thrust component)

The second half depends on angle, and the angle depends on both how the wings are mounted and the front lift/rotation force of the machine (as well as weight and distance from pivot of course).

[Suzaku]

This topic is missing the most important ingreedient in regards to runway takeoff. The angle of attack for the wings relative to the direction the craft during takeoff.

[vexx32]

Hmm. If we want to factor in the angle of attack of the wings, it's gonna get too complicated, really. Especially for craft that have a natural angle of attack because of their landing gear configuration (since it'll be considerably harder to measure).

My point about controllable lift still stands, by the way, since if you're going to be building a craft without control surfaces, it is not going to handle like a plane at all anyway. Once they fix up the lift and drag models, ridiculous planes like EndlessWaves' will no longer be able to fly anywhere near as well, if at all.

I'm trying to build a logical equation that works, guys, not just now but preferably in future. Minor revisions, fine, but if you have to rework it every update, what good is it for newbies?

As to angle of attack, well... assuming that the angle of attack of your wings is either all the same, or you've taken an average of the angle of attack of all your wings (and control surfaces, if you've altered their angles as well), I'm thinking... perhaps...

(l*t)/(m*tan(a))

Probably not a perfect equation by any means... and having a 90 degree angle of attack will give an undefined. But then, a plane with a 90 degree angle of attack is either going straight up like a rocket (thus having a zero angle of attack, as it's angle relative to airflow, which is the reverse of the velocity vector for now in KSP, as it does not simulate wind) or not going anywhere... or just going to spin uncontrollably.

[NuclearWarfare]

Sometimes it depends. Some of my Cyclone aircraft don't take off till I leave the runway. Once it hits the end of the runway it will boost off the little peak from the runway to the ground.

[vexx32]

That usually means you just need to angle the craft in order for it to takeoff. There's a dropoff at the end of the runway, which gives you room to pitch up without the ground getting in the way. Try to alter the landing gear configuration and make the nose higher than the tail using just the landing gear. That'll help those situations. Also, canards on the nose can make it much easier to take off, but it makes it a little touchy to control.

[NuclearWarfare]

That usually means you just need to angle the craft in order for it to takeoff. There's a dropoff at the end of the runway, which gives you room to pitch up without the ground getting in the way. Try to alter the landing gear configuration and make the nose higher than the tail using just the landing gear. That'll help those situations. Also, canards on the nose can make it much easier to take off, but it makes it a little touchy to control.

No, no I must have worded that wrong. The planes do make it in the air, like the description you said Vexx

[vexx32]

Ah, yeah. Definitely want to make the front landing gear lower than the rest so as to give the effect of raising the nose on the runway.

[Guest butt head]

i know this is completely random and off topic but can i have the .craft file?

(i need some target practices:D)