TWR in orbit

[KerbMav]

I use a mod to see this, but it is a general question, so I will put it here.

In VAB I set KER to show me the dV for bodies without atmosphere for transfer stages, works fine.

But what puzzles me still after almost one year of playing is that the TWR on e.g. Gilly can be 45.8 or 3.5 on Ike for some vessels, but in open space it can be something like 0.2 ... why can I accellerate faster while on the surface of a body with gravity than high above in orbit?

[Guest]

There's no such thing as "TWR" in vacuum, since you're weighless. There can be TMR, which gives you acceleration in m/s, not Gs. It should be about 10 times less than Kerbin TWR, but for other, lower gravity bodies you can have a lot of TWR with a dinky TMR.

[Kasuha]

TWR does not make sense in orbit - TWR stands for Thrust to Weight Ratio and in orbit you have no weight.

When landed, you are not in orbit. If the planet below you disappeared (but the gravity stayed) you'd find yourself in a highly elliptic orbit with apoapsis at your initial position. You only stay motionless (relative to surface) because the surface is constantly pushing you upwards. TWR 1 exactly matches the force it is exerting on you and you need at least that to stop being landed and get your ship in orbit.

When you're in orbit, there's nothing pushing on you so you are free to accelerate as slow as you want.

[Streetwind]

KER will still show you TWR in orbit, for reference (you can use it to judge roughly how well you can accelerate with your currently active engines at your current vehicle mass).

That TWR is relative to the gravity you are experiencing. That is, the further away you get from a celestial body, the higher this TWR readout given by KER climbs. And if you change into the SoI of a different body with a different gravitational pull, the numbers will be completely different. Also, as you consume fuel this number keeps going up, since your mass goes down.

So yeah, it's perfectly possible for KER to show 0.2 in low Kerbin orbit, yet much higher on Gilly surface, all with the same engine configuration and vehicle mass.

[Padishar]

There's no such thing as "TWR" in vacuum, since you're weighless. There can be TMR, which gives you acceleration in m/s, not Gs. It should be about 10 times less than Kerbin TWR, but for other, lower gravity bodies you can have a lot of TWR with a dinky TMR.

Vacuum makes no difference to TWR. Weight is the force (measured in newtons) exerted on an object by gravity so every body that has mass in the entire universe (real or in KSP) has a weight because gravity affects everything with mass. In KSP it is only the body whose SOI you are in that exerts gravity.

TWR does not make sense in orbit - TWR stands for Thrust to Weight Ratio and in orbit you have no weight.

See above. When in orbit you do have weight because gravity is still acting on you or you wouldn't be in orbit. The weight force is less in orbit because gravity is inversely proportional to the square of the distance but it is never zero.

When landed, you are not in orbit. If the planet below you disappeared (but the gravity stayed) you'd find yourself in a highly elliptic orbit with apoapsis at your initial position. You only stay motionless (relative to surface) because the surface is constantly pushing you upwards. TWR 1 exactly matches the force it is exerting on you and you need at least that to stop being landed and get your ship in orbit.

Strictly speaking you need a TWR greater than 1 to stop being landed.

When you're in orbit, there's nothing pushing on you so you are free to accelerate as slow as you want.

Gravity is still pushing on you but you are going sideways fast enough to follow an elliptical path around the planet.

KER will still show you TWR in orbit, for reference (you can use it to judge roughly how well you can accelerate with your currently active engines at your current vehicle mass).

That TWR is relative to the gravity you are experiencing. That is, the further away you get from a celestial body, the higher this TWR readout given by KER climbs. And if you change into the SoI of a different body with a different gravitational pull, the numbers will be completely different. Also, as you consume fuel this number keeps going up, since your mass goes down.

So yeah, it's perfectly possible for KER to show 0.2 in low Kerbin orbit, yet much higher on Gilly surface, all with the same engine configuration and vehicle mass.

Basically, yes. In the VAB/SPH, KER shows the TWR using the surface gravity of the selected reference body. In flight, it displays 3 different TWR values:

TWR (Throttle) - Ratio between the thrust your engines are currently generating and your current weight force

TWR (Current) - Ratio between the maximum thrust your engines can generate and your current weight force

TWR (Surface) - Ratio between the maximum thrust of engines and your weight force on the surface of the body whose SOI you are in

[Kasuha]

See above. When in orbit you do have weight because gravity is still acting on you or you wouldn't be in orbit. The weight force is less in orbit because gravity is inversely proportional to the square of the distance but it is never zero.

Okay, there are two slightly different views at the matter.

First of all, when you're in orbit, you definitely have (inertial) mass. That mass is the same regardless of what body are you orbiting, be it Kerbin, Eve, or Gilly.

Second, you don't have any measurable weight. Any weight measuring device brought to orbit will measure zero. I am omitting some effects here such as tidal forces or atmospheric drag in Earth orbits. The only device that helps you there is a device measuring your inertial mass. That device will measure the same value regardless which body you're orbiting.

And third, whether you do or don't have weight depends on whether you assume centrifugal and centripetal forces are real or not.

When you come from assumption that they're real, you have weight which is compensated by centrifugal force. When you come from assumption they're not real forces, then you can't have weight.

Strictly speaking you need a TWR greater than 1 to stop being landed.

Yes

[Padishar]

Okay, there are two slightly different views at the matter.

First of all, when you're in orbit, you definitely have (inertial) mass. That mass is the same regardless of what body are you orbiting, be it Kerbin, Eve, or Gilly.

Second, you don't have any measurable weight. Any weight measuring device brought to orbit will measure zero. I am omitting some effects here such as tidal forces or atmospheric drag in Earth orbits. The only device that helps you there is a device measuring your inertial mass.

And third, whether you do or don't have weight depends on whether you assume centrifugal and centripetal forces are real or not.

When you come from assumption that they're real, you have weight which is compensated by centrifugal force. When you come from assumption they're not real forces, then you can't have weight.

Don't complicate the issue with centrifugal/centripetal forces. They are totally dependent on the reference frame and only serve to confuse the matter. The only force acting on a ship in orbit is the gravitational attraction accelerating the ship directly towards the centre of the body being orbited.

There are two (well, 4 if you count "ISO" and "apparent") definitions of weight in common use but only the gravitational definition makes any sense when talking about TWR. If you try to use the "operational definition" then a rocket which has 5kN of gravity acting on it and 3kN of engine thrust would have a weight of only 2kN and hence a TWR of 1.5 which is clearly "wrong" as the rocket does not lift off. The "gravitational definition" gives a weight of 5kN and a "correct" TWR of 0.6.

[KerbMav]

Basically, yes. In the VAB/SPH, KER shows the TWR using the surface gravity of the selected reference body. In flight, it displays 3 different TWR values:

TWR (Throttle) - Ratio between the thrust your engines are currently generating and your current weight force

TWR (Current) - Ratio between the maximum thrust your engines can generate and your current weight force

TWR (Surface) - Ratio between the maximum thrust of engines and your weight force on the surface of the body whose SOI you are in

So this explains something, at least my confusion. (Should have asked these things months ago. )

I guess the maximum/current acceleration is what you are using to calculate the remaining burn time next to the stage dV numbers?

Could you implement a readout for the acceleration in KER?

(And now it has become a modding thread after all! )

[Kasuha]

Don't complicate the issue with centrifugal/centripetal forces. They are totally dependent on the reference frame and only serve to confuse the matter. The only force acting on a ship in orbit is the gravitational attraction accelerating the ship directly towards the centre of the body being orbited.

There are two (well, 4 if you count "ISO" and "apparent") definitions of weight in common use but only the gravitational definition makes any sense when talking about TWR. If you try to use the "operational definition" then a rocket which has 5kN of gravity acting on it and 3kN of engine thrust would have a weight of only 2kN and hence a TWR of 1.5 which is clearly "wrong" as the rocket does not lift off. The "gravitational definition" gives a weight of 5kN and a "correct" TWR of 0.6.

You might notice that weight is dependent on reference frame as well. It makes sense only if you use reference frame of the planet on which you're sitting.

I have no problems with definition of TWR of a rocket sitting on surface (of a non-rotating planet) but I still have to find any sense in TWR defined for rocket in motion without direct contact with surface.

Particularly for purposes of ascent to orbit and atmospheric pass, you need your rocket engine to exert force equivalent to 1 G to keep the ship at terminal velocity, but at the same time to compensate ineffectiveness of your ongoing maneuver. Statements like "you need TWR=2" don't make sense when you're at 30 km altitude, flying at 30 degrees to the surface and your speed is 1 km/s. There's no reasonable value you can multiply by two to get optimal thrust you need at that point.

And that's what I wanted to point out.

[Padishar]

I guess the maximum/current acceleration is what you are using to calculate the remaining burn time next to the stage dV numbers?

No, the burn time is calculated during the simulation of the engine burning to calculate the deltaV. It depends on the fuel consumption of the engines and how much fuel is in the stage. If the engines are throttled up then it uses the current throttle fuel consumption and when not running it should use the max throttle fuel consumption (but this may not be correct at the moment).

Could you implement a readout for the acceleration in KER?

There is the G-Force readout in the surface (SUR) tab. That is read directly from the Vessel object and should be useful...

[KerbMav]

Understood.

Ah, silly number two on my part - never thought about using the g-meter for this, thank you!

[technicalfool]

Don't complicate the issue with centrifugal/centripetal forces. They are totally dependent on the reference frame and only serve to confuse the matter. The only force acting on a ship in orbit is the gravitational attraction accelerating the ship directly towards the centre of the body being orbited.

And the force keeping that ship up in orbit and counteracting gravity is...?

Anyway, as others have said, the TWR that Flight Engineer, Mechjeb and similar give you, is based on the body you're either orbiting around or sat on (or tumbling uncontrollably toward).

[allmhuran]

And the force keeping that ship up in orbit and counteracting gravity is...?

Nothing. If there was an equal and opposite balancing force, then the ship could just hover in space motionless. It's not motionless, it's at orbital velocity. The ship is constantly falling due to gravity. The direction of the pull of gravity is constantly changing as the ship moves around the planet.

[Padishar]

I have no problems with definition of TWR of a rocket sitting on surface (of a non-rotating planet) but I still have to find any sense in TWR defined for rocket in motion without direct contact with surface.

Particularly for purposes of ascent to orbit and atmospheric pass, you need your rocket engine to exert force equivalent to 1 G to keep the ship at terminal velocity, but at the same time to compensate ineffectiveness of your ongoing maneuver. Statements like "you need TWR=2" don't make sense when you're at 30 km altitude, flying at 30 degrees to the surface and your speed is 1 km/s. There's no reasonable value you can multiply by two to get optimal thrust you need at that point.

And that's what I wanted to point out.

A TWR of 1 at any point (during flight or not) means that you can cancel out the downward acceleration due to gravity by burning straight up. A TWR of 2 means you can accelerate upwards at 1g. A TWR of 3 means you can get 2g. I agree that most of the statements like "you need TWR=2" aren't particularly accurate or helpful but it doesn't change the fact that, using the only definition of weight that makes TWR make any sense, your TWR is a well defined number no matter where you are, what direction you are travelling in or how fast you are going. You just need to think about the forces acting for the value to be useful.

[Redshift OTF]

It probably means you are orbiting the sun.

[technicalfool]

Nothing. If there was an equal and opposite balancing force, then the ship could just hover in space motionless. It's not motionless, it's at orbital velocity. The ship is constantly falling due to gravity. The direction of the pull of gravity is constantly changing as the ship moves around the planet.

This guy says it better than I can.

TL;DR: It may be an "apparent force", but you're still very really getting flung against the side of that fairground ride. Or being held against the pull of gravity. Either/or.

[allmhuran]

Yes, if you make your frame of reference rotate in the opposite direction of the orbit and at the same speed, and then draw the force diagram, then something will be wrong: The ship will continue straight ahead, even though there is only one force (gravity)acting on it! So you have to draw in an equal and opposite balancing force to account for the linear motion of the vessel. This equal and opposite balancing force is "centrifugal force", and we're allowed to add it precisely because we redrew the frame of reference.

But don't be fooled into thinking that this means there are actually two forces acting upon a vessel in orbit which add up to a net force of zero allowing it to "not fall". The net force acting on a vessel in orbit is non zero, that's what makes it change direction (F=ma, a change in direction is an acceleration). That non-zero net force is composed of gravity alone.

You're not really getting flung against the side of the fairground ride. You' body, the edge of the ride, and everything want so to go straight ("Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it") but the arms of the ride prevent the sides from going straight ahead (centripetal force), and the sides of the ride prevent you from going straight ahead. The ride is getting flung into you!

Edited May 24, 2014 by allmhuran