# Algorithms

## Simple motor feedforward

### Derivation

For a simple DC motor with the model

```
  dx/dt = −kᵥ/kₐ x + 1/kₐ u - kₛ/kₐ sgn(x),
```

where

```
  A = −kᵥ/kₐ
  B = 1/kₐ
  c = -kₛ/kₐ sgn(x)
  A_d = eᴬᵀ
  B_d = A⁻¹(eᴬᵀ - I)B
  dx/dt = Ax + Bu + c
```

Discretize the affine model.

```
  dx/dt = Ax + Bu + c
  dx/dt = Ax + B(u + B⁺c)
  xₖ₊₁ = eᴬᵀxₖ + A⁻¹(eᴬᵀ - I)B(uₖ + B⁺cₖ)
  xₖ₊₁ = A_d xₖ + B_d (uₖ + B⁺cₖ)
  xₖ₊₁ = A_d xₖ + B_d uₖ + B_d B⁺cₖ
```

Solve for uₖ.

```
  B_d uₖ = xₖ₊₁ − A_d xₖ − B_d B⁺cₖ
  uₖ = B_d⁺(xₖ₊₁ − A_d xₖ − B_d B⁺cₖ)
  uₖ = B_d⁺(xₖ₊₁ − A_d xₖ) − B⁺cₖ
```

Substitute in B assuming sgn(x) is a constant for the duration of the step.

```
  uₖ = B_d⁺(xₖ₊₁ − A_d xₖ) − kₐ(-(kₛ/kₐ sgn(x)))
  uₖ = B_d⁺(xₖ₊₁ − A_d xₖ) + kₐ(kₛ/kₐ sgn(x))
  uₖ = B_d⁺(xₖ₊₁ − A_d xₖ) + kₛ sgn(x)
```

Simplify the model when kₐ = 0.

Simplify A.

```
  A = −kᵥ/kₐ
```

As kₐ approaches zero, A approaches -∞.

```
  A = −∞
```

Simplify A_d.

```
  A_d = eᴬᵀ
  A_d = exp(−∞)
  A_d = 0
```

Simplify B_d.

```
  B_d = A⁻¹(eᴬᵀ - I)B
  B_d = A⁻¹((0) - I)B
  B_d = A⁻¹(-I)B
  B_d = -A⁻¹B
  B_d = -(−kᵥ/kₐ)⁻¹(1/kₐ)
  B_d = (kᵥ/kₐ)⁻¹(1/kₐ)
  B_d = kₐ/kᵥ(1/kₐ)
  B_d = 1/kᵥ
```

Substitute these into the feedforward equation.

```
  uₖ = B_d⁺(xₖ₊₁ − A_d xₖ) + kₛ sgn(x)
  uₖ = (1/kᵥ)⁺(xₖ₊₁ − (0) xₖ) + kₛ sgn(x)
  uₖ = (1/kᵥ)⁺(xₖ₊₁) + kₛ sgn(x)
  uₖ = kᵥxₖ₊₁ + kₛ sgn(x)
  uₖ = kₛ sgn(x) + kᵥxₖ₊₁
```

Simplify the model when kₐ ≠ 0

```
  uₖ = B_d⁺(xₖ₊₁ − A_d xₖ)
```

where

```
  A = −kᵥ/kₐ
  B = 1/kₐ
  A_d = eᴬᵀ
  B_d = A⁻¹(eᴬᵀ - I)B
```

When kᵥ = 0, A = 0 and B_d has a singularity. We can eliminate the singularity using the matrix exponential discretization method.

```
 [A  B]
 [0  0]T   [A_d  B_d]
e        = [ 0    I ]

 [0  B]
 [0  0]T   [1  BT]
e        = [0   1]

A_d = 1
B_d = BT
```

## Elevator feedforward

### Derivation

For an elevator with the model

```
  dx/dt = −kᵥ/kₐ x + 1/kₐ u - kg/kₐ - kₛ/kₐ sgn(x)
```

where

```
  A = −kᵥ/kₐ
  B = 1/kₐ
  c = -(kg/kₐ + kₛ/kₐ sgn(x))
  A_d = eᴬᵀ
  B_d = A⁻¹(eᴬᵀ - I)B
  dx/dt = Ax + Bu + c
```

Discretize the affine model.

```
  dx/dt = Ax + Bu + c
  dx/dt = Ax + B(u + B⁺c)
  xₖ₊₁ = eᴬᵀxₖ + A⁻¹(eᴬᵀ - I)B(uₖ + B⁺cₖ)
  xₖ₊₁ = A_d xₖ + B_d (uₖ + B⁺cₖ)
  xₖ₊₁ = A_d xₖ + B_d uₖ + B_d B⁺cₖ
```

Solve for uₖ.

```
  B_d uₖ = xₖ₊₁ − A_d xₖ − B_d B⁺cₖ
  uₖ = B_d⁺(xₖ₊₁ − A_d xₖ − B_d B⁺cₖ)
  uₖ = B_d⁺(xₖ₊₁ − A_d xₖ) − B⁺cₖ
```

Substitute in B assuming sgn(x) is a constant for the duration of the step.

```
  uₖ = B_d⁺(xₖ₊₁ − A_d xₖ) − kₐ(-(kg/kₐ + kₛ/kₐ sgn(x)))
  uₖ = B_d⁺(xₖ₊₁ − A_d xₖ) + kₐ(kg/kₐ + kₛ/kₐ sgn(x))
  uₖ = B_d⁺(xₖ₊₁ − A_d xₖ) + kg + kₛ sgn(x)
```

Simplify the model when kₐ = 0.

Simplify A.

```
  A = −kᵥ/kₐ
```

As kₐ approaches zero, A approaches -∞.

```
  A = −∞
```

Simplify A_d.

```
  A_d = eᴬᵀ
  A_d = exp(−∞)
  A_d = 0
```

Simplify B_d.

```
  B_d = A⁻¹(eᴬᵀ - I)B
  B_d = A⁻¹((0) - I)B
  B_d = A⁻¹(-I)B
  B_d = -A⁻¹B
  B_d = -(−kᵥ/kₐ)⁻¹(1/kₐ)
  B_d = (kᵥ/kₐ)⁻¹(1/kₐ)
  B_d = kₐ/kᵥ(1/kₐ)
  B_d = 1/kᵥ
```

Substitute these into the feedforward equation.

```
  uₖ = B_d⁺(xₖ₊₁ − A_d xₖ) + kg + kₛ sgn(x)
  uₖ = (1/kᵥ)⁺(xₖ₊₁ − (0) xₖ) + kg + kₛ sgn(x)
  uₖ = (1/kᵥ)⁺(xₖ₊₁) + kg + kₛ sgn(x)
  uₖ = kᵥxₖ₊₁ + kg + kₛ sgn(x)
  uₖ = kₛ sgn(x) + kg + kᵥxₖ₊₁
```

Simplify the model when kₐ ≠ 0

```
  uₖ = B_d⁺(xₖ₊₁ − A_d xₖ)
```

where

```
  A = −kᵥ/kₐ
  B = 1/kₐ
  A_d = eᴬᵀ
  B_d = A⁻¹(eᴬᵀ - I)B
```

When kᵥ = 0, A = 0 and B_d has a singularity. We can eliminate the singularity using the matrix exponential discretization method.

```
 [A  B]
 [0  0]T   [A_d  B_d]
e        = [ 0    I ]

 [0  B]
 [0  0]T   [1  BT]
e        = [0   1]

A_d = 1
B_d = BT
```

## Closed form Kalman gain for continuous Kalman filter with A = 0 and C = I

### Derivation

Model is

```
  dx/dt = Ax + Bu
  y = Cx + Du
```

where A = 0, B = 0, C = I, and D = 0.

The optimal cost-to-go is the P that satisfies

```
  AᵀP + PA − PBR⁻¹BᵀP + Q = 0
```

Let A = Aᵀ and B = Cᵀ for state observers.

```
  AP + PAᵀ − PCᵀR⁻¹CP + Q = 0
```

Let A = 0, C = I.

```
  −PR⁻¹P + Q = 0
```

Solve for P. P, Q, and R are all diagonal, so this can be solved element-wise.

```
  −pr⁻¹p + q = 0
  −pr⁻¹p = −q
  pr⁻¹p = q
  p²r⁻¹ = q
  p² = qr
  p = √(qr)
```

Now solve for the Kalman gain.

```
  K = PCᵀ(CPCᵀ + R)⁻¹
  K = P(P + R)⁻¹
  k = p(p + r)⁻¹
  k = √(qr)(√(qr) + r)⁻¹
  k = √(qr)/(√(qr) + r)
```

Multiply by √(q/r)/√(q/r).

```
  k = q/(q + r √(q/r))
  k = q/(q + √(qr²/r))
  k = q/(q + √(qr))
```

### Corner cases

For q = 0 and r ≠ 0,

```
  k = 0/(0 + √0)
  k = 0/0
```

Apply L'Hôpital's rule to k with respect to q.

```
  k = 1/(1 + r/(2 √(qr)))
  k = 2 √(qr)/(2 √(qr) + r)
  k = 2 √(0)/(2 √0 + r)
  k = 0/r
  k = 0
```

For q ≠ 0 and r = 0,

```
  k = q / (q + √0)
  k = q / q
  k = 1
```

## Quaternion to Euler angle conversion

### Conventions

We'll use the extrinsic X-Y-Z rotation order for Euler angles. The direction of rotation is CCW looking into the positive axis. If you point your right thumb along the positive axis direction, your fingers curl in the direction of rotation.

The angles are a\_x around the X-axis, a\_y around the Y-axis, and a\_z around the Z-axis, with the following constraints:

```
  -π ≤ a_x ≤ π
  -π/2 ≤ a_y ≤ π/2
  -π ≤ a_z ≤ π
```

The coordinate system is right-handed. If you point your right thumb along the +Z axis, your fingers curl from the +X axis to the +Y axis.

The quaternion imaginary numbers are defined as follows:

```
  îĵ = k̂
  ĵk̂ = î
  k̂î = ĵ
  îĵ = -k̂
  k̂ĵ = -î
  îk̂ = -ĵ
  î² = ĵ² = k̂² = -1
```

### Quaternion representation of axis rotations

We will take it as given that a rotation by θ radians around a normalized vector v is represented with the quaternion cos(θ/2) + sin(θ/2) (v\_x î + v\_y ĵ + v\_z k̂).

### Derivation

For convenience, we'll define the following variables:

```
  c_x = cos(a_x/2)
  s_x = sin(a_x/2)
  c_y = cos(a_y/2)
  s_y = sin(a_y/2)
  c_z = cos(a_z/2)
  s_z = sin(a_z/2)
```

We can calculate the quaternion corresponding to a set of Euler angles by applying each rotation in sequence. Recall that quaternions are composed with left multiplication, like matrices.

```
  q = (cos(a_z/2) + sin(a_z/2) k̂)(cos(a_y/2) + sin(a_y/2) ĵ)(cos(a_x/2) + sin(a_x/2) î)
  q = (c_z + s_z k̂)(c_y + s_y ĵ)(c_x + s_x î)
  q = (c_y c_z - s_y s_z î + s_y c_z ĵ + c_y s_z k̂)(c_x + s_x î)
    = (c_x c_y c_z + s_x s_y s_z)
      + (s_x c_y c_z - c_x s_y s_z) î
      + (c_x s_y c_z + s_x c_y s_z) ĵ
      + (c_x c_y s_z - s_x s_y c_z) k̂
```

Letting q = q\_w + q\_x î + q\_y ĵ + q\_z k̂, we can extract the components of the quaternion:

```
  q_w = c_x c_y c_z + s_x s_y s_z
  q_x = s_x c_y c_z - c_x s_y s_z
  q_y = c_x s_y c_z + s_x c_y s_z
  q_z = c_x c_y s_z - s_x s_y c_z
```

### Solving for a\_y

Solving for sin(a\_y):

```
  sin(a_y) = 2 c_y s_y
  sin(a_y) = 2 (c_x² c_y s_y + s_x² c_y s_y)
  sin(a_y) = 2 (c_x² c_y s_y c_z² + c_x² c_y s_y s_z²
              + s_x² c_y s_y c_z² + s_x² c_y s_y s_z²)
  sin(a_y) = 2 (c_x² c_y s_y c_z² + s_x² c_y s_y s_z²
              + s_x² c_y s_y c_z² + c_x² c_y s_y s_z²)
  sin(a_y) = 2 (c_x² c_y s_y c_z² + c_x s_x c_y² c_z s_z
              + c_x s_x s_y² c_z s_z + s_x² c_y s_y s_z²
              - c_x s_x c_y² c_z s_z + s_x² c_y s_y c_z²
              + c_x² c_y s_y s_z² - c_x s_x s_y² c_z s_z)
  sin(a_y) = 2 ((c_x c_y c_z + s_x s_y s_z)(c_x s_y c_z + s_x c_y s_z)
              - (s_x c_y c_z - c_x s_y s_z)(c_x c_y s_z - s_x s_y c_z))
  sin(a_y) = 2 (q_w q_y - q_x q_z)
```

Then solving for a\_y:

```
  a_y = sin⁻¹(sin(a_y))
  a_y = sin⁻¹(2 (q_w q_y - q_x q_z))
```

### Solving for a\_x and a\_z

Solving for cos(a\_x) cos(a\_y):

```
  cos(a_x) cos(a_y) = (cos²(a_x/2) - sin²(a_x/2))(cos²(a_y/2) - sin²(a_y/2))
  cos(a_x) cos(a_y) = (c_x² - s_x²)(c_y² - s_y²)
  cos(a_x) cos(a_y) = c_x² c_y² - c_x² s_y² - s_x² c_y² + s_x² s_y²
  cos(a_x) cos(a_y) = c_x² (1 - s_y²) - c_x² s_y² - s_x² c_y² + s_x² (1 - c_y²)
  cos(a_x) cos(a_y) = c_x² - c_x² s_y² - c_x² s_y² - s_x² c_y² + s_x² - s_x² c_y²
  cos(a_x) cos(a_y) = c_x² + s_x² - 2 (c_x² s_y² + s_x² c_y²)
  cos(a_x) cos(a_y) = 1 - 2 (c_x² s_y² + s_x² c_y²)
  cos(a_x) cos(a_y) = 1 - 2 (c_x² s_y² c_z² + c_x² s_y² s_z²
                           + s_x² c_y² c_z² + s_x² c_y² s_z²)
  cos(a_x) cos(a_y) = 1 - 2 (s_x² c_y² c_z² + c_x² s_y² s_z²
                           + c_x² s_y² c_z² + s_x² c_y² s_z²)
  cos(a_x) cos(a_y) = 1 - 2 (s_x² c_y² c_z² - 2 c_x s_x c_y s_y c_z s_z + c_x² s_y² s_z²
                           + c_x² s_y² c_z² + 2 c_x s_x c_y s_y c_z s_z + s_x² c_y² s_z²)
  cos(a_x) cos(a_y) = 1 - 2 ((s_x c_y c_z - c_x s_y s_z)² + (c_x s_y c_z + s_x c_y s_z)²)
  cos(a_x) cos(a_y) = 1 - 2 (q_x² + q_y²)
```

Solving for sin(a\_x) cos(a\_y):

```
  sin(a_x) cos(a_y) = (2 cos(a_x/2) sin(a_x/2))(cos²(a_y/2) - sin²(a_y/2))
  sin(a_x) cos(a_y) = (2 c_x s_x)(c_y² - s_y²)
  sin(a_x) cos(a_y) = 2 (c_x s_x c_y² - c_x s_x s_y²)
  sin(a_x) cos(a_y) = 2 (c_x s_x c_y² c_z² + c_x s_x c_y² s_z²
                       - c_x s_x s_y² c_z² - c_x s_x s_y² s_z²)
  sin(a_x) cos(a_y) = 2 (c_s s_x c_y² c_z² - c_x s_x s_y² s_z²
                       - c_x s_x s_y² c_z² + c_x s_x c_y² s_z²)
  sin(a_x) cos(a_y) = 2 (c_x s_x c_y² c_z² - c_x² c_y s_y c_z s_z
                       + s_x² c_y s_y c_z s_z - c_x s_x s_y² s_z²
                       + c_x² c_y s_y c_z s_z - c_x s_x s_y² c_z²
                       + c_x s_x c_y² s_z² - s_x² c_y s_y c_z s_z)
  sin(a_x) cos(a_y) = 2 ((c_x c_y c_z + s_x s_y s_z)(s_x c_y c_z - c_x s_y s_z)
                       + (c_x s_y c_z + s_x c_y s_z)(c_x c_y s_z - s_x s_y c_z))
  sin(a_x) cos(a_y) = 2 (q_w q_x + q_y q_z)
```

Similarly, solving for cos(a\_z) cos(a\_y):

```
  cos(a_z) cos(a_y) = (cos²(a_z/2) - sin²(a_z/2))(cos²(a_y/2) - sin²(a_y/2))
  cos(a_z) cos(a_y) = (c_z² - s_z²)(c_y² - s_y²)
  cos(a_z) cos(a_y) = c_y² c_z² - s_y² c_z² - c_y² s_z² + s_y² s_z²
  cos(a_z) cos(a_y) = c_y² (1 - s_z²) - s_y² c_z² - c_y² s_z² + s_y² (1 - c_z²)
  cos(a_z) cos(a_y) = c_y² - c_y² s_z² - s_y² c_z² - c_y² s_z² + s_y² - s_y² c_z²
  cos(a_z) cos(a_y) = c_y² + s_y² - 2 (c_y² s_z² + s_y² c_z²)
  cos(a_z) cos(a_y) = 1 - 2 (c_y² s_z² + s_y² c_z²)
  cos(a_z) cos(a_y) = 1 - 2 (c_x² c_y² s_z² + s_x² c_y² s_z²
                           + c_x² s_y² c_z² + s_x² s_y² c_z²)
  cos(a_z) cos(a_y) = 1 - 2 (c_x² s_y² c_z² + s_x² c_y² s_z²
                           + c_x² c_y² s_z² + s_x² s_y² c_z²)
  cos(a_z) cos(a_y) = 1 - 2 (c_x² s_y² c_z² + 2 c_x s_x c_y s_y c_z s_z + s_x² c_y² s_z²
                           + c_x² c_y² s_z² - 2 c_x s_x c_y s_y c_z s_z + s_x² s_y² c_z²)
  cos(a_z) cos(a_y) = 1 - 2 ((c_x s_y c_z + s_x c_y s_z)² + (c_x c_y s_z - s_x s_y c_z)²)
  cos(a_z) cos(a_y) = 1 - 2 (q_y² + q_z²)
```

Similarly, solving for sin(a\_z) cos(a\_y):

```
  sin(a_z) cos(a_y) = (2 cos(a_z/2) sin(a_z/2))(cos²(a_y/2) - sin²(a_y/2))
  sin(a_z) cos(a_y) = (2 c_z s_z)(c_y² - s_y²)
  sin(a_z) cos(a_y) = 2 (c_y² c_z s_z - s_y² c_z s_z)
  sin(a_z) cos(a_y) = 2 (c_x² c_y² c_z s_z + s_x² c_y² c_z s_z
                       - c_x² s_y² c_z s_z - s_x² s_y² c_z s_z)
  sin(a_z) cos(a_y) = 2 (c_x² c_y² c_z s_z - s_x² s_y² c_z s_z
                       + s_x² c_y² c_z s_z - c_x² s_y² c_z s_z)
  sin(a_z) cos(a_y) = 2 (c_x² c_y² c_z s_z - c_x s_x c_y s_y c_z²
                       + c_x s_x c_y s_y s_z² - s_x² s_y² c_z s_z
                       + c_x s_x c_y s_y c_z² + s_x² c_y² c_z s_z
                       - c_x² s_y² c_z s_z - c_x s_x c_y s_y s_z²)
  sin(a_z) cos(a_y) = 2 ((c_x c_y c_z + s_x s_y s_z)(c_x c_y s_z - s_x s_y c_z)
                       + (s_x c_y c_z - c_x s_y s_z)(c_x s_y c_z + s_x c_y s_z))
  sin(a_z) cos(a_y) = 2 (q_w q_z + q_x q_y)
```

Solving for a\_x and a\_z:

```
  a_x = atan2(sin(a_x), cos(a_x))
  a_z = atan2(sin(a_z), cos(a_z))
```

If cos(a\_y) > 0:

```
  a_x = atan2(sin(a_x) cos(a_y), cos(a_x) cos(a_y))
  a_z = atan2(sin(a_z) cos(a_y), cos(a_z) cos(a_y))
  a_x = atan2(2 (q_w q_x + q_y q_z), 1 - 2 (q_x² + q_y²))
  a_z = atan2(2 (q_w q_z + q_x q_y), 1 - 2 (q_y² + q_z²))
```

Because -π/2 ≤ a\_y ≤ π/2, cos(a\_y) ≥ 0. Therefore, the only remaining case is cos(a\_y) = 0, whose only solutions in that range are a\_y = ±π/2.

```
  a_y = ±π/2
  a_y/2 = ±π/4
  cos(a_y/2) = √2/2
  c_y = √2/2
  sin(a_y/2) = ±√2/2
  s_y = ±√2/2
```

Plugging into the quaternion components:

```
  q_w = c_x c_y c_z + s_x s_y s_z
  q_x = s_x c_y c_z - c_x s_y s_z
  q_y = c_x s_y c_z + s_x c_y s_z
  q_z = c_x c_y s_z - s_x s_y c_z
  q_w = √2/2 c_x c_z ± √2/2 s_x s_z
  q_x = √2/2 s_x c_z ∓ √2/2 c_x s_z
  q_y = ±√2/2 c_x c_z + √2/2 s_x s_z
  q_z = √2/2 c_x s_z ∓ √2/2 s_x c_z
  q_w = √2/2 (c_x c_z ± s_x s_z)
  q_x = √2/2 (s_x c_z ∓ c_x s_z)
  q_y = √2/2 (± c_x c_z + s_x s_z)
  q_z = √2/2 (c_x s_z ∓ s_x c_z)
  q_w = √2/2 cos(a_z/2 ∓ a_x/2)
  q_x = √2/2 sin(a_x/2 ∓ a_z/2)
  q_y = √2/2 -cos(a_x/2 ∓ a_z/2)
  q_z = √2/2 sin(a_z/2 ∓ a_x/2)
```

In either case only the sum or the difference between a\_x and a\_z can be determined. We'll pick the solution where a\_x = 0.

```
  q_w = √2/2 cos(a_z/2 ∓ 0)
  q_w = √2/2 cos(a_z/2)
  cos(a_z/2) = √2 q_w
  q_z = √2/2 sin(a_z/2 ∓ 0)
  q_z = √2/2 sin(a_z/2)
  sin(a_z/2) = √2 q_z
  cos(a_z) = cos²(a_z/2) - sin²(a_z/2)
  cos(a_z) = (√2 q_w)² - (√2 q_z)²
  cos(a_z) = 2 q_w² - 2 q_z²
  cos(a_z) = 2 (q_w² - q_z²)
  sin(a_z) = 2 cos(a_z/2) sin(a_z/2)
  sin(a_z) = 2 (√2 q_w) (√2 q_z)
  sin(a_z) = 4 q_w q_z
  a_z = atan2(4 q_w q_z, 2 (q_w² - q_z²))
  a_z = atan2(2 q_w q_z, q_w² - q_z²)
```

### Determining if cos(a\_y) ≈ 0

When calculating a\_x:

```
  cos(a_y) ≈ 0
  cos²(a_y) ≈ 0
  cos²(a_x) cos²(a_y) + sin²(a_x) cos²(a_y) ≈ 0
  (cos(a_x) cos(a_y))² + (sin(a_x) cos(a_y))² ≈ 0
```

Note that this reuses the cos(a\_x) cos(a\_y) and sin(a\_x) cos(a\_y) terms needed to calculate a\_x.

When calculating a\_z:

```
  cos(a_y) ≈ 0
  cos²(a_y) ≈ 0
  cos²(a_y) cos²(a_z) + cos²(a_y) sin²(a_z) ≈ 0
  (cos(a_y) cos(a_z))² + (cos(a_y) sin(a_z))² ≈ 0
```

Note that this reuses the cos(a\_y) cos(a\_z) and cos(a\_y) sin(a\_z) terms needed to calculate a\_z.

## Quaternion Exponential

We will take it as given that a quaternion has scalar and vector components `𝑞 = s + 𝑣⃗`, with vector component 𝑣⃗ consisting of a unit vector and magnitude `𝑣⃗ = θ v̂`.

```
𝑞 = s + 𝑣⃗

𝑣⃗ = θv̂

exp(𝑞) = exp(s + 𝑣⃗)
exp(𝑞) = exp(s)exp(𝑣⃗)
exp(𝑞) = exp(s)exp(θ v̂)
```

Applying euler's identity:

```
exp(θv̂) = cos(θ) + sin(θ) v̂
```

Gives us:
```
exp(𝑞) = exp(s) [cos(θ) + sin(θ) v̂]
```

Rearranging `𝑣⃗ = θ v̂` we can solve for v̂: `v̂ = 𝑣⃗ / θ`

```
exp(𝑞) = exp(s) [cos(θ) + 𝑣⃗ sin(θ) / θ]
```

## Quaternion Logarithm

We will take it as a given that for a given quaternion of the form `𝑞 = s + 𝑣⃗`, we can calculate the exponential: `exp(𝑞) = exp(s) [cos(θ) + 𝑣⃗ sin(θ) / θ]` where `θ = ||𝑣⃗||`.

Additionally, `exp(log(𝑞)) = q` for a given value of `log(𝑞)`. There are multiple solutions to `log(𝑞)` caused by the imaginary axes in 𝑣⃗, discussed here: https://en.wikipedia.org/wiki/Complex_logarithm

We will demonstrate the principal solution of `log(𝑞)` satisfying `exp(log(𝑞)) = q`.
This being `log(𝑞) = log(||𝑞||) + 𝑣⃗ atan2(θ, s) / θ`, is the principal solution to `log(𝑞)` because the function `atan2(θ, s)` returns the principal value corresponding to its arguments.

Proof: `log(𝑞) = log(||𝑞||) + 𝑣⃗ atan2(θ, s) / θ` satisfies `exp(log(𝑞)) = q`.

```
exp(log(𝑞)) = exp(log(||𝑞||) + 𝑣⃗ atan2(θ, s) / θ)


exp(log(𝑞)) = exp(log(||𝑞||)) exp(𝑣⃗ atan2(θ, s) / θ)

Substitutions:
𝑣⃗ = θ v̂:
exp(log(||𝑞||)) = ||𝑞||
exp(log(𝑞)) = ||𝑞|| exp(atan2(θ, s) v̂)

exp(log(𝑞)) = ||𝑞|| [cos(atan2(θ, s)) + sin(atan2(θ, s)) v̂]

Substitutions:
cos(atan2(θ, s)) = s / √(θ² + s²)
sin(atan2(θ, s)) = θ / √(θ² + s²)

exp(log(𝑞)) = ||𝑞|| [s / √(θ² + s²) + θ v̂ / √(θ² + s²)]

√(θ² + s²) = ||𝑞||

exp(log(𝑞)) = ||𝑞|| [s / ||𝑞|| + θ v̂ / ||𝑞||]
exp(log(𝑞)) = s + θ v̂

exp(log(𝑞)) = s + 𝑣⃗

exp(log(𝑞)) = 𝑞
```

## Unit Quaternion in SO(3) from Rotation Vector in 𝖘𝖔(3)

We will take it as a given that members of 𝖘𝖔(3) take the form `𝑣⃗ = θ v̂`, representing a rotation θ around a unit axis v̂.

We additionally take it as a given that quaternions in SO(3) are of the form `𝑞 = cos(θ / 2) + sin(θ / 2) v̂`, representing a rotation of θ around unit axis v̂.

```
θ = ||𝑣⃗||
v̂ = 𝑣⃗ / θ

𝑞 = cos(θ / 2) + sin(θ / 2) v̂
𝑞 = cos(||𝑣⃗|| / 2) + sin(||𝑣⃗|| / 2) 𝑣⃗ / ||𝑣⃗||
```

## Rotation vector in 𝖘𝖔(3) from Unit Quaternion in SO(3)

We will take it as a given that members of 𝖘𝖔(3) take the form  `𝑟⃗ = θ r̂`, representing a rotation θ around a unit axis r̂.

We additionally take it as a given that quaternions in SO(3) are of the form `𝑞 = s + 𝑣⃗ = cos(θ / 2) + sin(θ / 2) v̂`, representing a rotation of θ around unit axis v̂.

```
s + 𝑣⃗ = cos(θ / 2) + sin(θ / 2) v̂
s = cos(θ / 2)
𝑣⃗ = sin(θ / 2) v̂
||𝑣⃗|| = sin(θ / 2)

θ / 2 = atan2(||𝑣⃗||, s)
θ = 2 atan2(||𝑣⃗||, s)

r̂ = 𝑣⃗ / ||𝑣⃗||

𝑟⃗ = θr̂
𝑟⃗ = 2 atan2(||𝑣⃗||, s) 𝑣⃗ / ||𝑣⃗||
```

## Closed form solution for an Exponential Motion Profile

### [Derivation of continuous-time model](wpimath/algorithms/ExponentialProfileModel.py)


### Heuristic for input direction in Exponential Profile

Demonstration: https://www.desmos.com/calculator/3jamollwrk

The fastest path possible for an exponential profile (and the placement of the inflection point) depend on boundary conditions.

Specifically, the placement (xf, vf) relative to the possible trajectories that cross through (x0, v0) decides this. There are two possible trajectories to take from the initial state. In the desmos demo these are colored Green and Purple, which arise from applying +input and -input from the initial state respectively. Red and Yellow trajectories arise from applying -input and +input respectively from terminal conditions.

In order to reach the terminal state from the initial state by following Green in the +v direction, the second step is following Red in the -v direction.
Likewise, Purple must be followed in the -v direction, and then Yellow must be followed in the +v direction.

The specific conditions surrounding this decision are fourfold:
- A: v0 >= 0
- B: vf >= 0
- C: vf >= x1_ps(vf, U)
- D: vf >= x1_ps(vf, -U)

Where x1_ps(v, U) follows the Green line, and x1_ps(v, -U) follows the Purple line.

This creates a decision table:
| v0>=0 | vf>=0 | vf>=x1_ps(vf,U) | vf>=x1_ps(vf,-U) | Output Sign |
|-------|-------|-----------------|------------------|------------:|
| False | False | False           | False            |          -1 |
| False | False | False           | True             |           1 |
| False | False | True            | False            |           1 |
| False | False | True            | True             |           1 |
| False | True  | False           | False            |          -1 |
| False | True  | False           | True             |          -1 |
| False | True  | True            | False            |           1 |
| False | True  | True            | True             |           1 |
| True  | False | False           | False            |          -1 |
| True  | False | False           | True             |           1 |
| True  | False | True            | False            |          -1 |
| True  | False | True            | True             |           1 |
| True  | True  | False           | False            |          -1 |
| True  | True  | False           | True             |          -1 |
| True  | True  | True            | False            |          -1 |
| True  | True  | True            | True             |           1 |

Which is equivalent to `-1 if (A & ~D) | (B & ~C) | (~C & ~D) else 1`.
