[T.V.C] part 01) Vector Mechanics for Rotating Frame

INTRODUCTION

 

In part00, I introduced concept of ' TVC ' short of ' Thrust Vector Control ' and Prerequisites like ' Attitude Dynamics ' ,' PID Control ' and the other things.

So in this part, we're going to learn about ' Attitude Dynamics' and widen the scope to 'DCM' (Direction Cosine Matrix), Euler Angle and Quaternions in next part. These 3 methods have different merits and demerits, so you can choose the most advantageous one.

 

Basic Vector Mechanics for Rotating Frame

In Particle

** Notation : Bold letter indicates Vector and dot(  ̇) indicates differential element.

First, Let's find out basic vector mechanics for Rotating Frame.

Let's see the figure below. There are ' Inertia Frame ' {n1, n2, n3} and ' Body Frame ' {b1, b2, b3}. Inertia Frame is The coordinate that has no external forces, so particles stop or move with constant velocity.

R = X n1 + Y n2 + Z n3
R0 = X0 n1 + Y0 n2 + Z0 n3
r = x b1 + y b2 + z b3

Therefore Distance vector would be R = R0 + r. Velocity is derived from differential R.

R ̇= R0 ̇+ ṙ.

R0 ̇ = X0 ̇ n1 + Y0 ̇ n2 + Z0 ̇ n3
ṙ  = x ̇ b1 + y ̇ b2 + z ̇ b3 + x b1 ̇  + y b2 ̇  + z b3 ̇ You might curious about why ṙ has differentiated frame components {b1 ̇, b2 ̇, b3 ̇} but R0 ̇ hasn't. Because ' Inertia Frame ' doesn't change but ' Body Frame ' does depends on time, so n1 ̇ , n2 ̇ and n3 ̇ are 0.


So we need to find out the differential of Body Frame, {b1 ̇, b2 ̇, b3 ̇}. But first, don't forget that Body do both Translation and Rotation. So time derivative of Body Frame vector is function of Angular Velocity vector.

By Differential Formula,

According to the figure above, Δbi is part of circumference which bi moves. So | Δbi | = | bi | * sin(⌀) * Δθ.
(Length of Circumference) = (Radius) * (Central Angle)

We suppose the angle Δθ is very small, so Δbi has same direction with tangential direction of bi, et.

Hence Δbi = ( | bi |*sin(⌀)*Δθ ) et. Consequently bi ̇ = lim(Δt->0) Δbi / Δt = lim(Δt->0) ( | bi |*sin(⌀)*Δθ ) et / Δt = ( | bi |*sin(⌀) ) lim(Δt->0) (Δθ/Δt) et = | bi |*sin(⌀)*| ω | et = ω X bi (Cross Product's result can easily find by right hand rule).

Then apply above bi ̇ = ω X et to ṙ ,
ṙ = x ̇ b1 + y ̇ b2 + z ̇ b3 + x b1 ̇ + y b2 ̇ + z b3 ̇
  = x ̇ b1 + y ̇ b2 + z ̇ b3 + x(ω X b1) + y(ω X b2) + z(ω X b3)
  = x ̇ b1 + y ̇ b2 + z ̇ b3 + ω X (x b1 + y b2 + z b3)
= rb ̇ + ω X r

Hence R ̇= R0 ̇+ ṙ = R0 ̇ + rb ̇+ ω X r .


But we will deal with forces applied on Body, so one more differential of R ̇ is needed. You can easily get the result if you refer to upper process.

R ̇̇  = R0 ̇̇ + ẋ̇  b1 + ẏ̇  b2 + ż̇  b3 + x ̇ b1 ̇ + y ̇ b2 ̇ + z ̇ b3 ̇ + ω ̇ X r + ω X ṙ
= R0 ̇̇ + rb ̇̇ + ω X rb ̇ + ω ̇X r + ω X ( rb ̇ + ω X r )
= R0 ̇̇  + rb ̇̇  + 2ω X rb ̇ + ω ̇X r + ω X ( ω X r )

In this equation, 2ω X rb ̇ called ' Coriolis's Acceleration '. Coriolis's Force is one of Inertia forces applied on moving object in Rotation Frame. It acts on tangential direction of rotation axis like Centrifugal Force.


So we learned about Acceleration vector on Body Frame. Next let's find out ' Linear Momentum ' and ' Angular Momentum '.

First, Linear Momentum ' G ' is vector who indicates momentum in translational motion. G = mV, Ġ = ma = ∑F

Second, Angular Momentum ' H ' is vector who indicates momentum in rotational motion. H = r X G = r X (mV).

Ḣ = ṙ X mV + r X mV̇ (Differential of Cross product has same process) = V X mV + r X mV̇ = r X mV̇ (Cross product between two same vectors is zero) = r X ma = r X F = ∑M (Moment).

Consequently, G = mV, Ġ = ma = ∑F , H = r X (mV), Ḣ = ∑M = T(Torque).

 

Representative of Rigid Body

So let's go to next stage.

C : The center of mass
P : an arbitrary point fixed in the body

By definition of 'Center of mass', mRc = ∫∫∫b R dm (b stands for body) so Rc = ∫∫∫ R dm/m. By Newton's 2nd Law(F=ma), mRċ ̇ = F.

You already knew that there's an attraction between particles even though they have very small mass, but this doesn't include that force and other reaction forces. Then follows Angular Momentum equation above, Hn = ∫∫∫b r X (dr/dt) dm = ∫∫∫b r X ṙ dm.
And differential of Angular Momentum is Hṅ = ∫∫∫b ṙ  X ṙ dm + ∫∫∫b r X r ̇̇ dm = ∫∫∫b r X r ̇̇ dm = ∫∫∫b r X (R ̇̇ - Rp ̇̇ ) dm = Tp + Rp ̇̇ X ∫∫∫b r dm (Tp is applied Torque and A X B = - B X A).

 

EULER'S ROTATIONAL EQUATION

Next let's figure the Euler's Rotational equation of motion for rigid body.

O : The center of mass in rigid body

Angular Momentum at point O is Ho = mr X ṙ = ∫ r X ( ω X r ) dm. (V = ṙ = ω X r ).
Since r = x b1 + y b2 + z b3, ω = ωx b1 + ωy b2 + ωz b3, apply vectors to equation above,
r X ( ω X r ) = { ωx(y^2 + z^2) - ωy(xy) - ωz(xz) } b1
+ { -ωx(xy) + ωy(x^2 + z^2) - ωz(yz) } b2
+ { -ωx(xz) - ωy(yz) + ωz(x^2 + y^2) } b3

We can change this equation to shorten one by ' Moment of Inertia ' . Moment of Inertia is magnitude of property that an object who rotates on the axis of rotation to sustain rotation and denoted by ' I '. (Check 'Moment of Inerita' part https://loookup.tistory.com/6) So the magnitude for each axis is
Ix = ∫b (y^2 + z^2) dm, Iy = ∫b (x^2 + z^2) dm, Iz = ∫b (x^2 + y^2) dm
Ixy = ∫b (xy) dm, Iyz = ∫b (yz) dm, Ixz = ∫b (xz) dm

 

Hence, Ho = ∫ r X ( ω X r ) dm = { ωx(Ix) - ωy(Ixy) - ωz(Ixz) } b1
+ { -ωx(Ixy) + ωy(Iy) - ωz(Iyz) } b2
+ { -ωx(Ixz) - ωy(Iyz) + ωz(Iz) } b3

Since ω = {ωx, ωy, ωz}, we can change Moment of Inertia to Matrix form. Which is

Thus, Ho = Iω.


Then according to result above, H = hx b1 + hy b2 + hz b3 and Ḣ = hẋ b1 + hẏ b2 + hż b3 + hx b1̇ + hy b2̇ + hz b3̇ = hḃ + ω X h = Iω̇ + ω X Iω = T.


Finally we get Euler's rotational equation of motion for rigid body.

Where ωx is skew-symmetric matrix

At last, we learned about basic vector mechanics for Rotating Frame and Rotational equation by Angular Momentum and Inertial Matrix. I'm worried about confusing notation and also counter-intuitive formula.

I hope you can easily understand what I said and step forward to next part.


See you at next part !!

 

 

Reference : Space Science, 2022-1, Donghyun Cho

 

 

Thanks for your reading and I hope you got what you wanted !

 

Thanks for people who give me Inspiration and Source.