For an entirely random reason, I am wondering about thrust vectoring on spacecraft. More precisely, how do I calculate the angular acceleration imparted on a ship by an off-axis thrust vector? (assuming the off-center burn is only displaced in one direction in a two-dimensional plane, to not complicate things)

I am reasonably sure my attempts to do that are wrong: I am essentially modelling the situation in two dimensions, by splitting the off-axis thrust vector into its component forces, taking the component that acts perpedincular to the direction of travel and applying it as I would a lever, with the arm being the engine's distance from the center of mass.

I'm not even sure this is high school level physics ; At least my old high school textbooks don't have a problem like this one. Maybe Polish science education just sucks, I don't know

## Thrust vectoring and math

**Moderators:** Alyrium Denryle, SCRawl, Thanas

### Thrust vectoring and math

**JULY 20TH 1969**- The day the entire world was looking up

*It suddenly struck me that that tiny pea, pretty and blue, was the Earth. I put up my thumb and shut one eye, and my thumb blotted out the planet Earth. I didn't feel like a giant. I felt very, very small.*

**- NEIL ARMSTRONG, MISSION COMMANDER, APOLLO 11**

Signature dedicated to the greatest achievement of mankind.

MILDLY DERANGED PHYSICIST does not mind BREAKING the SOUND BARRIER, because it is INSURED. - Simon_Jester considering the problems of hypersonic flight for Team L.A.M.E.

### Re: Thrust vectoring and math

You need to know the "moment of inertia" of your ship.

Forces act in both a rotational sense and a linear sense. Your force will both accelerate linearly as normal, but will also provide a rotational acceleration about the centre of mass.

In linear terms you have F = m a

In rotational mechanics there is a simple analogue

C = I q

Where C is the torque: force times perpendicular distance between line of action of the force and the centre of mass (force applied in the rotational sense),

I is moment of inertia and

q is angular acceleration (in radians/second^2 or whatever units you have)

The moment of inertia of an object is calculated by adding up the masses of all the component objects times the distance from the centre of mass squared m r^2

Forces act in both a rotational sense and a linear sense. Your force will both accelerate linearly as normal, but will also provide a rotational acceleration about the centre of mass.

In linear terms you have F = m a

In rotational mechanics there is a simple analogue

C = I q

Where C is the torque: force times perpendicular distance between line of action of the force and the centre of mass (force applied in the rotational sense),

I is moment of inertia and

q is angular acceleration (in radians/second^2 or whatever units you have)

The moment of inertia of an object is calculated by adding up the masses of all the component objects times the distance from the centre of mass squared m r^2

Apparently nobody can see you without a signature.

### Re: Thrust vectoring and math

So if the engine is, say, 10 metres from the centre of mass, and the spacecraft masses 15 tonnes, the moment of inertia is 15 000 * 100. At least I got that correct

So I'm just wondering how to arrive from:

Force of thrust

Deflection angle

At the torque applied towards rotating the spacecraft.

So I'm just wondering how to arrive from:

Force of thrust

Deflection angle

At the torque applied towards rotating the spacecraft.

**JULY 20TH 1969**- The day the entire world was looking up

*It suddenly struck me that that tiny pea, pretty and blue, was the Earth. I put up my thumb and shut one eye, and my thumb blotted out the planet Earth. I didn't feel like a giant. I felt very, very small.*

**- NEIL ARMSTRONG, MISSION COMMANDER, APOLLO 11**

Signature dedicated to the greatest achievement of mankind.

MILDLY DERANGED PHYSICIST does not mind BREAKING the SOUND BARRIER, because it is INSURED. - Simon_Jester considering the problems of hypersonic flight for Team L.A.M.E.

### Re: Thrust vectoring and math

Sort my shitty phone just submitted before I was done and I was too late with my edit so I Lost all the rest of the typing.

Treating the object as a point mass is unfortunately not valid.

In your example the axis of rotation is through the centre of mass as the ship is not being clamped anywhere. This means the moment of inertia is entirety dependant on the shape of the ship, so you need much more detail on that.

Torque is the easy bit. Imagine a line that your engine thrusts along. Find the perpendicular distance between that line and the centre of mass and then multiply that distance by the force your engine puts out and you have the torque. For example, if he engine thrusts directly towards or away from the centre of mass the torque is zero. If the engine is 10m away from the com and points perpendicular to he line between them then the distance is just 10m and the torque is thrust * 10m.

To illustrate how the geometry of he ship matters, imagine you are in a flying saucer. The ship is a disc of mass m and radius r and has uniform density. If we treat it as a point mass then I about the com would be 0. If however we treat it as a series of particles and add up all the little m*r*r contributions we get I=1/2 m r^2.

Treating the object as a point mass is unfortunately not valid.

In your example the axis of rotation is through the centre of mass as the ship is not being clamped anywhere. This means the moment of inertia is entirety dependant on the shape of the ship, so you need much more detail on that.

Torque is the easy bit. Imagine a line that your engine thrusts along. Find the perpendicular distance between that line and the centre of mass and then multiply that distance by the force your engine puts out and you have the torque. For example, if he engine thrusts directly towards or away from the centre of mass the torque is zero. If the engine is 10m away from the com and points perpendicular to he line between them then the distance is just 10m and the torque is thrust * 10m.

To illustrate how the geometry of he ship matters, imagine you are in a flying saucer. The ship is a disc of mass m and radius r and has uniform density. If we treat it as a point mass then I about the com would be 0. If however we treat it as a series of particles and add up all the little m*r*r contributions we get I=1/2 m r^2.

Apparently nobody can see you without a signature.

### Re: Thrust vectoring and math

Yeah, just like Steel explains, moment of inertia is related to the integral of the mass of the object over its dimensions. An object with weight and length has a larger moment of inertia if the weight is in two iron balls on either end of a long rod, versus one iron ball in the center of that long rod. (Think Discovery from 2001 Space Odyssey vs. a spherical ship of the same mass).

Anyway, you're right that you need to decompose the thrust into its components. Take the line between your engine and the center of mass; the components of thrust that are perpendicular to this will create torque, and the component of thrust along this line will add velocity. The torque is equal to the force times the distance to the center of mass. In vector math, the torque is equal to the cross product of the force and distance to center of mass. Then, angular acceleration uses moment of inertia and the formula Steel gave.

So the steps are

1) Find the direction/distance between the engine and center of mass

2) Decompose the thrust into perpendicular and parallel to this line

3) Torque C is the perpendicular component times length of this line

4) a) Find moment of inertia I (involves integrals, but standard formulas exist for standard shapes)

4) b) Rotational acceleration q = C / I

Anyway, you're right that you need to decompose the thrust into its components. Take the line between your engine and the center of mass; the components of thrust that are perpendicular to this will create torque, and the component of thrust along this line will add velocity. The torque is equal to the force times the distance to the center of mass. In vector math, the torque is equal to the cross product of the force and distance to center of mass. Then, angular acceleration uses moment of inertia and the formula Steel gave.

So the steps are

1) Find the direction/distance between the engine and center of mass

2) Decompose the thrust into perpendicular and parallel to this line

3) Torque C is the perpendicular component times length of this line

4) a) Find moment of inertia I (involves integrals, but standard formulas exist for standard shapes)

4) b) Rotational acceleration q = C / I

### Re: Thrust vectoring and math

Yeah, high school level physics. God, do I suck at this after barely ten years out of school.

Thanks for the help, dudes

Thanks for the help, dudes

**JULY 20TH 1969**- The day the entire world was looking up

*It suddenly struck me that that tiny pea, pretty and blue, was the Earth. I put up my thumb and shut one eye, and my thumb blotted out the planet Earth. I didn't feel like a giant. I felt very, very small.*

**- NEIL ARMSTRONG, MISSION COMMANDER, APOLLO 11**

Signature dedicated to the greatest achievement of mankind.

MILDLY DERANGED PHYSICIST does not mind BREAKING the SOUND BARRIER, because it is INSURED. - Simon_Jester considering the problems of hypersonic flight for Team L.A.M.E.

### Re: Thrust vectoring and math

PeZook wrote:Yeah, high school level physics. God, do I suck at this after barely ten years out of school.

Thanks for the help, dudes

Well this bit for simple shapes at the basic level is 'high school', but if you want to get into more complicated stuff you'll need some degree level techniques.

Apparently nobody can see you without a signature.

### Re: Thrust vectoring and math

The point is that I couldn't even do it for a spherical chicken in vacuum without help.

**JULY 20TH 1969**- The day the entire world was looking up

**- NEIL ARMSTRONG, MISSION COMMANDER, APOLLO 11**

Signature dedicated to the greatest achievement of mankind.

MILDLY DERANGED PHYSICIST does not mind BREAKING the SOUND BARRIER, because it is INSURED. - Simon_Jester considering the problems of hypersonic flight for Team L.A.M.E.

### Re: Thrust vectoring and math

Hey, I wasn't taught this stuff in high school. I was only taught this stuff in my first and second years of undergrad engineering. Granted, I probably could have understood it in HS, but I don't think I knew enough about vector math to really get it back then.

### Re: Thrust vectoring and math

Oh, my high school maths program was just fine. I could do vectors, matrices and integral calculus pretty well. We did six hours of math a week in my senior year.

Come to think of it, the physics course might've been good as well and I just didn't pay much attention. This theory is supported by my physics grades...

Come to think of it, the physics course might've been good as well and I just didn't pay much attention. This theory is supported by my physics grades...

**JULY 20TH 1969**- The day the entire world was looking up

**- NEIL ARMSTRONG, MISSION COMMANDER, APOLLO 11**

Signature dedicated to the greatest achievement of mankind.

MILDLY DERANGED PHYSICIST does not mind BREAKING the SOUND BARRIER, because it is INSURED. - Simon_Jester considering the problems of hypersonic flight for Team L.A.M.E.

Return to “Science, Logic, And Morality”

### Who is online

Users browsing this forum: Yahoo [Bot] and 1 guest