Teacher PD. One table, run top to bottom, from a target point to the number you type into the arm. Do these yourself before you teach Week 3; that is the whole point of the session.
This arm, these numbers. L₁ = 18 cm, L₂ = 22 cm. Practical reach 8 to 40 cm. Shoulder travel 20° to 150°. Every solve in this unit is elbow-up, at both endpoints, which is why θ₁ = α + β throughout and never α − β.
The pipeline
Three gates, shaded, and you stop at any one that fails. The last row is the one most sheets stop short of: the elbow field on the arm is offset from the angle you solved, so θ₂′ = 144.4° is typed as -54.4° at the end point. Type the raw 144.4 instead and the tip lands 43.4 cm from the target, which is further than the arm is long.
α aims at the target. β lifts link 1 above that line. θ₁ = α + β. At the elbow, θ₂′ is the turn away from straight.
Worked: the motion's start point (18, 22)
Step
Equation
Value
L₁
given
18 cm
L₂
given
22 cm
x
target across
18
y
target up
22
r
r = √(x² + y²)
28.43 cm
Step 0 · reach check
is 8 ≤ r ≤ 40?
yes
cos θ₂′
(r² − L₁² − L₂²) ÷ (2 · L₁ · L₂)
0.000
in range?
is −1 ≤ cos θ₂′ ≤ 1?
yes
θ₂′ (elbow)
arccos(cos θ₂′)
90.0°
α
arctan(y ÷ x)
50.7°
cos β
(L₁² + r² − L₂²) ÷ (2 · L₁ · r)
0.633
β
arccos(cos β)
50.7°
θ₁ (shoulder, elbow-up)
α + β
101.4°
shoulder check
is 20° ≤ θ₁ ≤ 150°?
yes
elbow FIELD value
−θ₂′ + 90
0.0°
This is the canonical start point used all through Week 3, and it is worth knowing why it is convenient: r² = 18² + 22² = 808, which is exactly L₁² + L₂² = 324 + 484. That makes cos θ₂′ = 0 and θ₂′ exactly 90°, so the two links meet at a right angle.
Now the other end of the same motion
Same three angles, a target closer in. A bigger θ₂′ folds the arm and pulls the tip nearer the base.
Worked: the motion's end point (10, 8)
Step
Equation
Value
L₁
given
18 cm
L₂
given
22 cm
x
target across
10
y
target up
8
r
r = √(x² + y²)
12.81 cm
Step 0 · reach check
is 8 ≤ r ≤ 40?
yes
cos θ₂′
(r² − L₁² − L₂²) ÷ (2 · L₁ · L₂)
-0.813
in range?
is −1 ≤ cos θ₂′ ≤ 1?
yes
θ₂′ (elbow)
arccos(cos θ₂′)
144.4°
α
arctan(y ÷ x)
38.7°
cos β
(L₁² + r² − L₂²) ÷ (2 · L₁ · r)
0.009
β
arccos(cos β)
89.5°
θ₁ (shoulder, elbow-up)
α + β
128.2°
shoulder check
is 20° ≤ θ₁ ≤ 150°?
yes
elbow FIELD value
−θ₂′ + 90
-54.4°
The whole motion. Elbow-up at both ends, so the arm never flips shape mid-move.
The sweep. θ₁ goes 101.4° → 128.2° and θ₂′ goes 90.0° → 144.4°, elbow-up at both ends, then reverse. Both endpoints solved, one shape held: that is the Day 16 gate.
Your turn — four targets
Blank tables, same order. Answers overleaf. Work them by hand with the calculator in DEG.
1. Target (30, 10)
Step
Equation
Value
L₁
given
18 cm
L₂
given
22 cm
x
target across
y
target up
r
r = √(x² + y²)
Step 0 · reach check
is 8 ≤ r ≤ 40?
cos θ₂′
(r² − L₁² − L₂²) ÷ (2 · L₁ · L₂)
in range?
is −1 ≤ cos θ₂′ ≤ 1?
θ₂′ (elbow)
arccos(cos θ₂′)
α
arctan(y ÷ x)
cos β
(L₁² + r² − L₂²) ÷ (2 · L₁ · r)
β
arccos(cos β)
θ₁ (shoulder, elbow-up)
α + β
shoulder check
is 20° ≤ θ₁ ≤ 150°?
elbow FIELD value
−θ₂′ + 90
2. Target (20, 26)
Step
Equation
Value
L₁
given
18 cm
L₂
given
22 cm
x
target across
y
target up
r
r = √(x² + y²)
Step 0 · reach check
is 8 ≤ r ≤ 40?
cos θ₂′
(r² − L₁² − L₂²) ÷ (2 · L₁ · L₂)
in range?
is −1 ≤ cos θ₂′ ≤ 1?
θ₂′ (elbow)
arccos(cos θ₂′)
α
arctan(y ÷ x)
cos β
(L₁² + r² − L₂²) ÷ (2 · L₁ · r)
β
arccos(cos β)
θ₁ (shoulder, elbow-up)
α + β
shoulder check
is 20° ≤ θ₁ ≤ 150°?
elbow FIELD value
−θ₂′ + 90
3. Target (6, 30)
Step
Equation
Value
L₁
given
18 cm
L₂
given
22 cm
x
target across
y
target up
r
r = √(x² + y²)
Step 0 · reach check
is 8 ≤ r ≤ 40?
cos θ₂′
(r² − L₁² − L₂²) ÷ (2 · L₁ · L₂)
in range?
is −1 ≤ cos θ₂′ ≤ 1?
θ₂′ (elbow)
arccos(cos θ₂′)
α
arctan(y ÷ x)
cos β
(L₁² + r² − L₂²) ÷ (2 · L₁ · r)
β
arccos(cos β)
θ₁ (shoulder, elbow-up)
α + β
shoulder check
is 20° ≤ θ₁ ≤ 150°?
elbow FIELD value
−θ₂′ + 90
4. Target (34, 14)
Step
Equation
Value
L₁
given
18 cm
L₂
given
22 cm
x
target across
y
target up
r
r = √(x² + y²)
Step 0 · reach check
is 8 ≤ r ≤ 40?
cos θ₂′
(r² − L₁² − L₂²) ÷ (2 · L₁ · L₂)
in range?
is −1 ≤ cos θ₂′ ≤ 1?
θ₂′ (elbow)
arccos(cos θ₂′)
α
arctan(y ÷ x)
cos β
(L₁² + r² − L₂²) ÷ (2 · L₁ · r)
β
arccos(cos β)
θ₁ (shoulder, elbow-up)
α + β
shoulder check
is 20° ≤ θ₁ ≤ 150°?
elbow FIELD value
−θ₂′ + 90
Three that do not run
The shaded ring is every point the arm can reach: 8 to 40 cm. Red points sit outside it. The orange point is inside the ring and still impossible, which is why the shoulder check exists.
Each fails at a different gate, and only one of them fails in a way the algebra alone will catch. Say which gate stops each, and what you would tell the student.
A. Target (5, 4)
Step
Equation
Value
L₁
given
18 cm
L₂
given
22 cm
x
target across
y
target up
r
r = √(x² + y²)
Step 0 · reach check
is 8 ≤ r ≤ 40?
cos θ₂′
(r² − L₁² − L₂²) ÷ (2 · L₁ · L₂)
in range?
is −1 ≤ cos θ₂′ ≤ 1?
θ₂′ (elbow)
arccos(cos θ₂′)
α
arctan(y ÷ x)
cos β
(L₁² + r² − L₂²) ÷ (2 · L₁ · r)
β
arccos(cos β)
θ₁ (shoulder, elbow-up)
α + β
shoulder check
is 20° ≤ θ₁ ≤ 150°?
elbow FIELD value
−θ₂′ + 90
B. Target (30, 30)
Step
Equation
Value
L₁
given
18 cm
L₂
given
22 cm
x
target across
y
target up
r
r = √(x² + y²)
Step 0 · reach check
is 8 ≤ r ≤ 40?
cos θ₂′
(r² − L₁² − L₂²) ÷ (2 · L₁ · L₂)
in range?
is −1 ≤ cos θ₂′ ≤ 1?
θ₂′ (elbow)
arccos(cos θ₂′)
α
arctan(y ÷ x)
cos β
(L₁² + r² − L₂²) ÷ (2 · L₁ · r)
β
arccos(cos β)
θ₁ (shoulder, elbow-up)
α + β
shoulder check
is 20° ≤ θ₁ ≤ 150°?
elbow FIELD value
−θ₂′ + 90
C. Target (3, 14)
Step
Equation
Value
L₁
given
18 cm
L₂
given
22 cm
x
target across
y
target up
r
r = √(x² + y²)
Step 0 · reach check
is 8 ≤ r ≤ 40?
cos θ₂′
(r² − L₁² − L₂²) ÷ (2 · L₁ · L₂)
in range?
is −1 ≤ cos θ₂′ ≤ 1?
θ₂′ (elbow)
arccos(cos θ₂′)
α
arctan(y ÷ x)
cos β
(L₁² + r² − L₂²) ÷ (2 · L₁ · r)
β
arccos(cos β)
θ₁ (shoulder, elbow-up)
α + β
shoulder check
is 20° ≤ θ₁ ≤ 150°?
elbow FIELD value
−θ₂′ + 90
Answer key
Target
r
cos θ₂′
θ₂′
α
β
θ₁
Field
Verdict
(18, 22)
28.43
0.000
90.0°
50.7°
50.7°
101.4°
0.0°
Runs as solved.
(10, 8)
12.81
-0.813
144.4°
38.7°
89.5°
128.2°
-54.4°
Runs as solved.
(30, 10)
31.62
0.242
76.0°
18.4°
42.5°
60.9°
14.0°
Runs as solved.
(20, 26)
32.80
0.338
70.2°
52.4°
39.1°
91.6°
19.8°
Runs as solved.
(6, 30)
30.59
0.162
80.7°
78.7°
45.2°
123.9°
9.3°
Runs as solved.
(34, 14)
36.77
0.687
46.6°
22.4°
25.8°
48.2°
43.4°
Runs as solved.
(5, 4)
6.40
-0.968
165.6°
38.7°
121.1°
159.7°
-75.6°
The math solves, but r = 6.40 cm is inside the 8 cm floor. The elbow only closes to 160°, so the arm cannot fold this tight. Move the point.
(30, 30)
42.43
1.253
no solution — cos θ₂′ outside −1…1
Unreachable. r = 42.43 cm is past the 40 cm the arm can reach, and cos θ₂′ = 1.253 is outside −1…1, so no angle exists.
(3, 14)
14.32
-0.761
139.6°
77.9°
85.0°
162.9°
-49.6°
Reachable, but θ₁ = 162.9° is outside the shoulder's 20°–150°. Correct mathematics the arm cannot hold. Move the point or re-aim the base.
What each failure teaches
(5, 4) is the one that slips through. r = 6.40 cm clears the geometric floor of |L₂ − L₁| = 4 cm, and cos θ₂′ stays inside −1…1, so the algebra produces an answer. The arm still cannot do it, because the elbow stops closing at 160°. This is why students get 8 cm and not 4 cm.
(30, 30) fails honestly. cos θ₂′ = 1.253, outside −1…1, so arccos has no value and the calculator errors. The error is the answer.
(3, 14) is reachable and still impossible. θ₁ = 162.9° is past the shoulder's 150°. Correct mathematics the machine cannot hold, which is the hardest of the three to spot because nothing on paper looks wrong.
Four things to hold constant
The links are 18 and 22, always. Every solve is elbow-up, so θ₁ = α + β and never α − β, which is what keeps one shape across the motion. The reach floor is 8 cm, from the elbow stop, not the geometric 4 cm. And β comes from the Law of Cosines, cos β = (L₁² + r² − L₂²) ÷ (2 · L₁ · r), which returns a positive magnitude and so has no sign to get wrong.