Intuition What this page is for
The parent note showed you the machinery. Here we stress-test it: every angle range, every sign, the identity, the singularity edge, a real spacecraft word problem, and an exam trap. If you can follow all nine examples below, no MRP scenario can surprise you.
This page is self-contained: the two boxed facts below are all you strictly need, though the parent note derives them from scratch.
Before symbols fly:
e ^ (read "e-hat") is a length-1 arrow pointing along the rotation axis — the pole you spin around.
Φ (capital phi) is how far you spin. Convention on this page: Φ is in degrees , unless a formula explicitly says "in radians" (only the small-angle approximation in Example 6 does).
tan is "opposite over adjacent" of a right triangle; here it turns the angle Φ/4 into a number , and that number is the whole magnitude of σ .
Definition Two symbols we will lean on
q 0 — the scalar part of the quaternion of the same rotation, q 0 = cos ( Φ/2 ) . It runs from + 1 (no rotation) down to − 1 (a full turn). The MRP blow-up happens exactly when q 0 = − 1 ; we compute q 0 explicitly in Example 6 to locate that singularity.
[ σ × ] — the cross-product matrix (a "skew-symmetric" matrix) built from σ = ( σ x , σ y , σ z ) . It is the 3 × 3 table that turns "cross with σ " into ordinary matrix multiplication:
[ σ × ] = 0 σ z − σ y − σ z 0 σ x σ y − σ x 0 .
Its diagonal is all zeros and mirror entries are negatives of each other — that is what "skew-symmetric" means.
Every case this topic can throw at you falls into one of these cells. Each worked example below names the cell it covers.
Cell
What makes it special
Covered by
A. Small angle (Φ < 18 0 ∘ )
∥ σ ∥ < 1 , no switch needed
Ex 1
B. Exactly 18 0 ∘
∥ σ ∥ = 1 exactly, the switch boundary
Ex 2
C. Large angle (18 0 ∘ < Φ < 36 0 ∘ )
∥ σ ∥ > 1 , must switch to shadow set
Ex 3
D. Zero / identity
Φ = 0 , σ = 0 — degenerate axis
Ex 4
E. Tilted axis, signed components
axis not along a coordinate line; negative entries
Ex 5
F. Limiting behaviour
Φ → 36 0 ∘ , blow-up; and Φ → 0 ∘
Ex 6
G. Propagation (kinematics)
use σ ˙ = 4 1 B ω off-identity
Ex 7
H. Real-world word problem
detumble → convert → decide switch
Ex 8
I. Exam twist
recover Φ , e ^ back from σ
Ex 9
Read the figure carefully — it is the map for every example:
The horizontal axis is the rotation angle Φ , in degrees, from 0 ∘ to about 37 5 ∘ .
The vertical axis is the MRP magnitude ∥ σ ∥ = tan ( Φ/4 ) .
The chalk-blue curve is that tangent. It starts flat near the origin (small angles give small σ ), then curves upward ever faster.
The pale-yellow dashed line at height 1 is the switch threshold; the yellow arrow marks where the curve crosses it, exactly at Φ = 18 0 ∘ .
The pink dotted vertical line at Φ = 36 0 ∘ is the singularity, and the pink arrow shows the curve racing to infinity as it approaches.
The shaded blue band under the curve, left of the yellow line, is the safe region where ∥ σ ∥ ≤ 1 .
The whole game is: stay in the shaded region , and if you ever drift out, jump to the shadow set.
Worked example Example 1 — Cell A: small angle, no switch
Rotate 9 0 ∘ about e ^ = ( 0 , 0 , 1 ) . Find σ .
Forecast: guess whether ∥ σ ∥ is above or below 1 before computing.
Magnitude: ∥ σ ∥ = tan ( Φ/4 ) = tan ( 22. 5 ∘ ) ≈ 0.4142 .
Why this step? The length of σ is always the quarter-angle tangent — that is the definition, nothing else to decide.
Direction: σ = e ^ ⋅ 0.4142 = ( 0 , 0 , 0.4142 ) .
Why this step? The axis e ^ points along z , so all the length lands on the z component.
Verify: ∥ σ ∥ = 0.4142 < 1 , so we are comfortably inside the safe region (Cell A confirmed). tan ( 22. 5 ∘ ) = 2 − 1 , a known value — matches. ✅
Worked example Example 2 — Cell B: exactly
18 0 ∘ , the boundary
Rotate 18 0 ∘ about e ^ = ( 1 , 0 , 0 ) . Find σ and say whether to switch.
Forecast: is 18 0 ∘ "safe" or "dangerous"? It sits exactly on the line.
Magnitude: ∥ σ ∥ = tan ( 18 0 ∘ /4 ) = tan ( 4 5 ∘ ) = 1 exactly.
Why this step? tan 4 5 ∘ = 1 is the fixed landmark — this is why 18 0 ∘ is the switch threshold.
Vector: σ = ( 1 , 0 , 0 ) .
Why this step? Length 1 along the x -axis.
Decision: at exactly ∥ σ ∥ = 1 the original and shadow set have equal magnitude (σ S = − σ /∥ σ ∥ 2 = ( − 1 , 0 , 0 ) , also length 1 ). Either branch is fine; convention is to switch once you exceed 1 .
Why this step? Shows the boundary is symmetric — neither branch is safer here.
Verify: shadow set σ S = ( − 1 , 0 , 0 ) has ∥ σ S ∥ = 1 = tan ( (( 360 − 180 ) /4 ) ∘ ) = tan 4 5 ∘ . Consistent. ✅
Worked example Example 3 — Cell C: large angle, must switch
Attitude is Φ = 27 0 ∘ about e ^ = ( 0 , 1 , 0 ) . Get the safe MRP set.
Forecast: predict the sign of the shadow-set component before doing the algebra.
Original magnitude: ∥ σ ∥ = tan ( 27 0 ∘ /4 ) = tan ( 67. 5 ∘ ) ≈ 2.4142 .
Why this step? Quarter-angle tangent again; 67. 5 ∘ is steep so the tangent is large.
Original vector: σ = ( 0 , 2.4142 , 0 ) . Since ∥ σ ∥ > 1 , we must switch.
Why this step? Being > 1 means we are past 18 0 ∘ heading toward the 36 0 ∘ blow-up.
Shadow set: σ S = − σ /∥ σ ∥ 2 = − ( 0 , 2.4142 , 0 ) / ( 2.4142 ) 2 = ( 0 , − 0.4142 , 0 ) .
Why this step? The shadow map flips sign and inverts magnitude — that is what − σ / ( σ ⊤ σ ) does.
Verify: ∥ σ S ∥ = 0.4142 = tan ( 22. 5 ∘ ) = tan ( (( 360 − 270 ) /4 ) ∘ ) . The shadow encodes Φ ′ = 9 0 ∘ , the "other way round" of 27 0 ∘ . Same physical attitude, now safely < 1 . ✅
Worked example Example 4 — Cell D: identity / zero rotation (degenerate axis)
No rotation at all: Φ = 0 ∘ . What is σ , and what about the axis?
Forecast: if there is no spin, is the axis even defined?
Magnitude: ∥ σ ∥ = tan ( 0/4 ) = tan 0 = 0 .
Why this step? Zero angle → zero quarter-angle → tangent is zero.
Vector: σ = 0 = ( 0 , 0 , 0 ) , regardless of what e ^ we "chose".
Why this step? Anything times 0 is 0 — this is why the axis is undefined at identity yet σ is still perfectly well-defined. MRPs smooth over the axis ambiguity that plagues axis–angle.
Verify: σ = 0 is the identity attitude, safely < 1 , and no singularity anywhere near. This is the clean start point used in Example 7. ✅
Worked example Example 5 — Cell E: tilted axis with signed components
Rotate 10 0 ∘ about the axis pointing to ( 1 , − 2 , 2 ) (not yet unit length). Find σ .
Forecast: the axis has a negative middle entry — will σ inherit that sign? (Yes.)
Normalise the axis. Length = 1 2 + ( − 2 ) 2 + 2 2 = 9 = 3 , so e ^ = ( 1/3 , − 2/3 , 2/3 ) .
Why this step? σ = e ^ tan ( Φ/4 ) demands a unit axis; skipping this scales your answer wrongly.
Magnitude: ∥ σ ∥ = tan ( 10 0 ∘ /4 ) = tan ( 2 5 ∘ ) ≈ 0.4663 .
Why this step? Length is quarter-angle tangent — independent of which way the axis tilts.
Vector: σ = 0.4663 ⋅ ( 1/3 , − 2/3 , 2/3 ) = ( 0.1554 , − 0.3108 , 0.3108 ) .
Why this step? Multiply the scalar length by each unit-axis component; the negative y carries straight through.
Verify: ∥ σ ∥ = 0.155 4 2 + 0.310 8 2 + 0.310 8 2 = 0.4663 = tan 2 5 ∘ < 1 . Direction recovers e ^ by dividing back out. ✅
Worked example Example 6 — Cell F: limiting behaviour near the singularity
Track ∥ σ ∥ as Φ → 36 0 ∘ and as Φ → 0 ∘ , and confirm where q 0 = − 1 .
Forecast: which limit blows up, which stays finite?
As Φ → 36 0 ∘ : Φ/4 → 9 0 ∘ , and tan ( 9 0 ∘ ) → + ∞ . Here the quaternion scalar q 0 = cos ( Φ/2 ) = cos ( 18 0 ∘ ) = − 1 .
Why this step? This is the lone MRP singularity, and computing q 0 pins it to exactly Φ = 36 0 ∘ . The switch rule exists precisely to keep you from reaching it.
Concrete near-value: at Φ = 35 8 ∘ , ∥ σ ∥ = tan ( 89. 5 ∘ ) ≈ 114.6 — enormous.
Why this step? Shows the blow-up is real and fast; even 2 ∘ short of 36 0 ∘ the numbers are already huge, so switch early .
As Φ → 0 ∘ : for a tiny angle the tangent is nearly the angle itself, so in radians tan ( Φ/4 ) ≈ Φ/4 ; hence σ → 0 .
Why this step? This small-angle rule only holds when the angle is measured in radians (the reason we flagged the degree convention at the top). It confirms the linearity used in kinematics, σ ˙ ≈ 4 1 ω .
Verify: tan ( 89. 5 ∘ ) = 114.588 … (finite but huge, matching the curve in the figure racing to ∞ ). The shadow of 35 8 ∘ is Φ ′ = 2 ∘ , ∥ σ S ∥ = tan ( 1 ∘ ) ≈ 0.01746 — tiny and safe. And cos ( 18 0 ∘ ) = − 1 confirms the singularity sits at Φ = 36 0 ∘ . ✅
Worked example Example 7 — Cell G: propagation off the identity
State σ = ( 0.1 , 0 , 0 ) , body rate ω = ( 0 , 0.4 , 0 ) rad/s. Find σ ˙ .
Forecast: the rate about y will feed mostly the y component of σ ˙ , but the σ x term twists in a cross-coupling. Guess which off-axis component appears.
Scalars: ∥ σ ∥ 2 = σ ⊤ σ = 0. 1 2 = 0.01 , so 1 − ∥ σ ∥ 2 = 0.99 .
Why this step? These feed the identity-scaling and outer-product blocks of B .
Build the cross-product matrix from the definition box, then B ( σ ) = ( 1 − ∥ σ ∥ 2 ) I 3 + 2 [ σ × ] + 2 σ σ ⊤ . With σ = ( 0.1 , 0 , 0 ) , [ σ × ] = 0 0 0 0 0 0.1 0 − 0.1 0 , so
B = 0.99 I 3 + 0 0 0 0 0 0.2 0 − 0.2 0 + 0.02 0 0 0 0 0 0 0 0 = 1.01 0 0 0 0.99 0.2 0 − 0.2 0.99 .
Why this step? 2 [ σ × ] contributes the ± 0.2 off-diagonals (that is the rotational cross-coupling); the outer product 2 σ σ ⊤ adds 2 ( 0.01 ) = 0.02 only to the xx slot. (I 3 is the 3 × 3 identity from the master-fact box.)
Apply: σ ˙ = 4 1 B ω = 4 1 1.01 0 0 0 0.99 0.2 0 − 0.2 0.99 0 0.4 0 = 4 1 0 0.396 0.08 = ( 0 , 0.099 , 0.02 ) .
Why this step? The 4 1 is the quarter-angle signature (not 2 1 ). Note the nonzero σ ˙ z = 0.02 — the predicted cross-coupling.
Verify: σ ˙ = ( 0 , 0.099 , 0.02 ) . As a sanity check, near identity (σ → 0 ) this would reduce to 4 1 ω = ( 0 , 0.1 , 0 ) — our 0.099 sits just below that, as expected for small nonzero σ . ✅
Worked example Example 8 — Cell H: real-world detumble word problem
A cubesat has just finished a slew and its attitude relative to target is Φ = 20 0 ∘ about e ^ = ( 0 , 0 , 1 ) . The flight software stores MRPs and switches whenever ∥ σ ∥ > 1 . What does it store, and does it switch?
Forecast: 20 0 ∘ is just past 18 0 ∘ — expect a switch, and expect the stored magnitude to be small , not large.
Raw MRP: ∥ σ ∥ = tan ( 20 0 ∘ /4 ) = tan ( 5 0 ∘ ) ≈ 1.1918 , so σ = ( 0 , 0 , 1.1918 ) .
Why this step? Convert attitude to MRP first; tan 5 0 ∘ > 1 flags the problem.
Trigger the switch since 1.1918 > 1 : σ S = − σ /∥ σ ∥ 2 = − ( 0 , 0 , 1.1918 ) / ( 1.1918 ) 2 = ( 0 , 0 , − 0.8391 ) .
Why this step? The controller needs bounded states; the shadow set keeps them away from the singularity so the control law stays well-conditioned.
Stored value: σ S = ( 0 , 0 , − 0.8391 ) , magnitude 0.8391 < 1 . Software stores this.
Why this step? This is the branch actually integrated by σ ˙ = 4 1 B ω going forward.
Verify: ∥ σ S ∥ = 0.8391 = tan ( 4 0 ∘ ) = tan ( (( 360 − 200 ) /4 ) ∘ ) , i.e. it encodes Φ ′ = 16 0 ∘ — the same orientation reached "the short way." Physically identical attitude, numerically safe. ✅
Worked example Example 9 — Cell I: exam twist (invert the map)
Given σ = ( 0 , 0.2679 , 0 ) , recover the rotation angle Φ and axis e ^ .
Forecast: which trig function undoes tan ? (arctan.) And how do we peel off the axis?
Magnitude first: ∥ σ ∥ = 0.2679 . Then Φ = 4 arctan ∥ σ ∥ .
Why this step? Since ∥ σ ∥ = tan ( Φ/4 ) , we ask "which quarter-angle has this tangent?" — that is exactly what arctan answers, and we multiply by 4 to undo the quarter.
Φ = 4 arctan ( 0.2679 ) = 4 × 1 5 ∘ = 6 0 ∘ .
Why this step? arctan ( 0.2679 ) = 1 5 ∘ is a clean landmark (tan 1 5 ∘ = 2 − 3 ≈ 0.2679 ).
Axis: e ^ = σ /∥ σ ∥ = ( 0 , 0.2679 , 0 ) /0.2679 = ( 0 , 1 , 0 ) .
Why this step? The unit direction of σ is the rotation axis; dividing by the length strips out the angle information cleanly.
Verify: rebuild σ = e ^ tan ( Φ/4 ) = ( 0 , 1 , 0 ) tan 1 5 ∘ = ( 0 , 0.2679 , 0 ) — matches the given input exactly. ✅
Recall One-line rules for every cell
Safe vs switch threshold?
∥ σ ∥ = 1 at Φ = 18 0 ∘ .
Undo the MRP map for the angle?
Φ = 4 arctan ∥ σ ∥ .
Undo the MRP map for the axis?
e ^ = σ /∥ σ ∥ .
Shadow set magnitude relation?
∥ σ S ∥ = tan ( (( 36 0 ∘ − Φ ) /4 ) ) .
Kinematic coefficient?
4 1 (quarter-angle), never 2 1 .
Related maps to compare against: Quaternions (Euler symmetric parameters) , Classical Rodrigues parameters (Gibbs vector) , Euler angles and Gimbal Lock , Direction Cosine Matrix (DCM) , and the propagation background in Attitude kinematics and $\boldsymbol\omega$ .