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Thrust vs Lift: Minmum TWR for spaceplanes?


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Hello KSP'ers!

So I was messing around on KSP the other day with this little guy (stock aero):

NzFtnSH.jpg

It's powered by two Ion Engines, and I made as much of it as possible with "massless" parts

So basically, I had KER open one time while flying it, and I realized that it was able to take off with a TWR of just 0.26(ish)

I'm not very knowlegable about aerodynamics or physics or anything, but I would assume a TWR of 1.0 is required to lift off. This seems to be the case with rockets, but not with planes (due to lift generating extra energy?)

My question is, consequently, two-fold:

1.) What is the lowest TWR anyone has taken off with?

2.) Is this still possible with "realistic" aero models? (NEAR, FAR, etc.)

Edited by Slam_Jones
Questions answered. Feel free to continue discussion if so desired :)
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My question is, consequently, two-fold:

1.) What is the lowest TWR anyone has taken off with?

2.) Is this still possible with "realistic" aero models? (NEAR, FAR, etc.)

1) I don't personally know about this... but as long as you have enough lift/weight you will eventually take off. TWR>1 is needed ONLY for rockets, since they start going up vertically. Planes usually have TWR<1 (unless they're extremely powerful ) and they exploit their velocity to generate lift ( IRL it's because pressure differences generated by wing shape, in KSP is just generated by the angle of attack, at least in FAR )

2) The exact craft you posted would probably never lift off or lift off and then crash in FAR. But this mod fortunately includes some very useful tools to verify if you plane will actually fly before bringing it on the runway

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1.) What is the lowest TWR anyone has taken off with?

2.) Is this still possible with "realistic" aero models? (NEAR, FAR, etc.)

1. I've taken off a 2 tonne aircraft with one ION engine. That makes my TWR 1 : 10 (or in decimal, 0.1). I could have kept adding wings until I weighed a lot more and I still would have taken off as long as overall thrust was higher than overall drag and high enough to increase speed such that overall lift higher than overall weight.

If this sounds very complicated, that's because compared to rocket science, it actually is.

TWR is basically the only thing you need to calculate to know if a rocket will fly, and the ratio of fuel mass to dry mass will then dictate maximum performance. With a plane, maximum performance depends on two other things besides TWR: aerodynamic lift and drag. Although drag can be a limiting factor for rockets (if the rocket is capable of exceeding terminal velocity), with a plane it is always a limiting factor, because you will spend a very long time flying through the thick atmosphere compared with a rocket that goes out of it by the most direct route - up.

2. I've never used NEAR or FAR but my gut feeling is yes.

Check out this thread for the recently-accomplished Ion Glider Collier Trophy Challenge for more examples of low TWR aircraft.

Edited by The_Rocketeer
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1. I've taken off a 2 tonne aircraft with one ION engine. That makes my TWR 1 : 1000. I could have kept adding wings until I weighed a lot more and I still would have taken off as long as overall thrust was higher than overall drag and high enough to increase speed such that overall lift higher than overall weight.

False. Your TWR was 0.1, since 2000 N / ( 2000 kg * 9.81 m/s^2 ) = 2000 N/19620 N= 0.102 ( on Earth/Kerbin )

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I'm not very knowlegable about aerodynamics or physics or anything, but I would assume a TWR of 1.0 is required to lift off. This seems to be the case with rockets, but not with planes (due to lift generating extra energy?)

The first law of thermodynamics means that nothing can generate or destroy energy, it just changes the kind of energy that you have from one type to another. This isn't strictly true in KSP (because KSP isn't real and the developers can basically build in cheats), it might help you to understand that the lift force is actually generated by part of the thrust force.

False. Your TWR was 0.1, since 2000 N / ( 2000 kg * 9.81 m/s^2 ) = 2000 N/19620 N= 0.102 ( on Earth/Kerbin )

You are in fact correct. I'd was working out 2000 N / 2000000 g. I've edited my last post.

Sorry not on top form. Been at a funeral today :-(

Edited by The_Rocketeer
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Thanks for the replies! My questions are basically answered, so I'll be changing the tag to "Answered." Feel free to keep discussing if you guys want, I love reading about stuff like this :)

Sorry not on top form. Been at a funeral today :-(

Sorry to hear that! Hope you feel better soon

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Basically the wings act as a thrust multiplier - instead of using your energy to accelerate a small amount of xenon to a very high velocity you can use it to accelerate a great amout of air to a much slower velocity, which is more efficient in terms of thrust to power ratio.

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Legitimately, I've put an ion glider into orbit with a .324 t/w ratio.

My jet/ rocket hybrid spaceplane SSTOs work at a peak t/w of 1.53.

Illegitimately, I've put a kraken drive infiniglider into orbit with a 0.00 t/w ratio.

Thrust to weight doesn't matter. What matters is lift- to- weight and thrust- to- drag.

Best,

-Slashy

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Excuse me while I put on my academic cap. The answers above give you the useful answers; what I give is the very limits of Kerbality.

Time to fire up the KSP-scripts.

First let's figure out what speed we need to be going for a lift surface to keep itself aloft:


import lift
import planet
def speed(AoA, surface):
return lift.speedForLift(surface.mass * planet.kerbin.gravity(), liftSurfaces = [(surface, AoA)])

speed(26, lift.wingStrake)
1.4777337523038174

I used 26 degrees for the AoA because that's the angle that gets you the best lift (approximately). Conclusion: we need to be going a bit shy of 1.5 m/s for a strake to lift itself. How much drag do we need to overcome?


def drag(AoA, surface):
return surface.dragForce(AoA, speed(AoA, surface))

drag(26, lift.wingStrake)
2.3416634787512635e-05

That's in kN; in whole numbers we're talking about 23 mN.

The thrust-to-weight ratio is therefore:


def twr(AoA, surface):
return drag(AoA, surface) / (surface.mass * planet.kerbin.gravity())

twr(26, lift.wingStrake)
9.548067191646333e-05

We conclude that 0.000095 is the minimal TWR you need to lift off using the wing strake placed at 26 degrees.

26 degrees is the best angle of attack for lift; is it for TWR?


for aoa in range(35):
print ("%2d: %.8f" % (aoa, twr(aoa, lift.wingStrake)))

Nope! 17 degrees is better: 0.00007966

We can do the same exercise with the delta-deluxe (but it has lift that increases all the way to 90 degrees).


for aoa in range(91):
print ("%2d: %.10f" % (aoa, twr(aoa, lift.deltaDeluxe)))

It maxes out at 90 degrees, 0.000023507 TWR.

These numbers mean that you can have a single ion engine power a plane with about 108k delta-deluxe wings, and it will lift off (yes, there's the mass of the ion engine, but it's 0.4t or so with xenon, whereas the wings are over 2,000t).

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The optimal AoA depends on a lot of things; the 17 degrees is for a strake that doesn't have to lift anything else. Maybe tonight if I have time I'll script up something more complete. The optimal AoA will depend on thrust, altitude, speed, and what your plane consists of.

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