Real-World Mathematics: Robotics, Kinematics, Coordinate Transforms and Path Planning

Application of Mathematics in Real-World Usage · Guide 33 · BTT Mathematics Hub

A robot never moves in “just one coordinate system.” Its camera has a frame, its arm links have frames, its base has a frame, and the room or map has a world frame. Robotics Mathematics is therefore largely the discipline of transforming positions, directions and velocities between frames without losing what each number means.

This guide uses ideal planar robots and fictional dimensions for Mathematics teaching. It is not engineering guidance for autonomous vehicles, industrial robots, drones, medical devices or other safety-critical machines. Real systems require calibrated sensors, dynamic limits, uncertainty models, collision certification and hardware-specific controls.

A point depends on the coordinate frame used

A point written as (2,1) means two coordinate units along one frame’s x-axis and one along its y-axis. If the axes rotate or the origin moves, the physical point can stay fixed while its coordinates change.

This is why robotics labels frames explicitly. A coordinate triple without a frame is incomplete information.

A planar rotation is a matrix transformation

For angle θ, the two-dimensional rotation matrix is R=[[cosθ,−sinθ],[sinθ,cosθ]].

At θ=90°, R maps (1,0) to (0,1). The point’s distance from the origin is preserved because rotation is rigid.

Translation changes origin without changing orientation

If a local point p=(1,2) is expressed in a frame whose origin lies at world position t=(4,3) with no rotation, its world position is p+t=(5,5).

Rotation and translation do different jobs, but robotics often needs them together.

Homogeneous coordinates combine rotation and translation

In 2D, a rigid transform can be written as a 3×3 homogeneous matrix T=[[R,t],[0,0,1]]. Multiplying T by [x,y,1]ᵀ applies both the rotation and translation.

If R is a 90° rotation and t=(4,3), local point (1,0) first rotates to (0,1) and then translates to world point (4,4).

Transform composition is order-sensitive

Rotate then translate is generally not the same as translate then rotate. Matrix multiplication reflects this non-commutativity.

For point (1,0), rotate 90° then add translation (4,0) to get (4,1). Translate first to (5,0), then rotate 90°, producing (0,5). Same ingredients, different order, different physical result.

The inverse transform returns coordinates to the earlier frame

For a rigid transform with rotation R and translation t, the inverse uses Rᵀ and translation −Rᵀt. Rotations are especially convenient because R⁻¹=Rᵀ.

A robust calculation should transform a test point forward and back and recover the original coordinates within numerical tolerance.

Forward kinematics maps joint variables to end-effector position

For a planar two-link arm with lengths l₁ and l₂ and joint angles θ₁ and θ₂, a standard model is x=l₁cosθ₁+l₂cos(θ₁+θ₂) and y=l₁sinθ₁+l₂sin(θ₁+θ₂).

Take l₁=2, l₂=1, θ₁=30° and θ₂=60°. Then θ₁+θ₂=90°, so x≈2(0.8660)+0=1.732 and y=2(0.5)+1=2.000.

Forward kinematics answers: given the joint configuration, where is the end effector?

Inverse kinematics reverses the question

Inverse kinematics asks which joint variables can reach a desired position. The answer can be unique, multiple, or nonexistent.

For a two-link planar arm, a target at radial distance r is reachable only when |l₁−l₂|≤r≤l₁+l₂. With lengths2 and1, reachable radial distances lie from1 to3 units.

The cosine rule exposes two-link inverse kinematics

For target radius r, cosθ₂=(r²−l₁²−l₂²)/(2l₁l₂). With l₁=l₂=1 and target (1,1), r²=2, so cosθ₂=0 and θ₂ can be ±90° under the two elbow branches.

The two mathematical solutions correspond to different physical configurations. A planner must choose one that also respects joint limits and obstacles.

The Jacobian maps joint rates to end-effector velocity

Differentiating the two-link equations gives a 2×2 Jacobian J relating [θ̇₁,θ̇₂]ᵀ to [ẋ,ẏ]ᵀ.

Its determinant simplifies to l₁l₂sinθ₂. When θ₂=0 orπ, the determinant is zero and the planar arm loses an instantaneous direction of motion: a kinematic singularity.

Differential-drive kinematics convert wheel speeds into body motion

For left and right linear wheel speeds vL and vR with wheel separation b, body forward speed is v=(vR+vL)/2 and angular speed is ω=(vR−vL)/b.

If vR=0.8m/s, vL=0.4m/s and b=0.5m, then v=0.6m/s and ω=0.8rad/s.

Equal wheel speeds produce straight-line motion in the ideal model

If both wheels move at0.5m/s, then ω=0 and the robot moves straight at0.5m/s. In four seconds it travels2m under the no-slip constant-speed assumption.

Unequal wheel speeds imply curvature

When ω≠0, ideal instantaneous turning radius of the robot centre is R=v/ω. For v=0.6 andω=0.8, R=0.75m.

This formula becomes unsuitable nearω=0 because the path tends toward a straight line and the radius tends toward infinity.

Path length is the sum of segment lengths

For waypoints (0,0)→(3,4)→(6,4), the first segment is5 and the second3, giving total path length8 units.

A shortest path depends on the allowed geometry. With obstacles, the Euclidean straight line may be impossible.

Obstacle inflation converts robot size into planning geometry

A circular robot of radius0.3m can be modelled as a point if every obstacle boundary is expanded outward by0.3m. This configuration-space idea moves geometry from the robot into the environment.

The method is exact only for the assumed robot shape and collision definition. Real robots may have orientation-dependent footprints.

A* search combines known cost and estimated remaining cost

A* ranks a search state with f(n)=g(n)+h(n), where g is cost already accumulated and h estimates cost to the goal.

If g=12 and an admissible heuristic h=5, f=17. A heuristic that never overestimates the remaining optimal cost supports optimality under the standard A* conditions.

Trajectory generation adds time to geometry

A geometric path says where to go. A trajectory says where to be at each time. One smooth scalar interpolation from0 to1 is q(s)=3s²−2s³ for0≤s≤1.

It satisfies q(0)=0, q(1)=1 and zero endpoint derivatives. At s=0.5, q=0.5.

Collision checking is a sampling problem when done discretely

If a planner checks only widely spaced points along a path, it can miss a narrow collision between samples. Finer sampling reduces this risk but increases computation.

Continuous collision methods or geometry-specific guarantees are needed when discrete samples cannot prove clearance.

Bearing to a target uses atan2, not an unsigned slope alone

From robot position (1,1) to target (4,5), displacement is(3,4). Distance is5 and bearing relative to the +x axis is atan2(4,3)≈53.13°.

atan2 preserves quadrant information that a simple arctangent of y/x can lose.

A complete robotics calculation states the frame and model assumptions

State which frame each quantity uses, joint conventions, link lengths, wheel geometry, collision footprint, path cost and whether dynamic effects are ignored.

The return path is configuration → transform → kinematics → feasible geometry → timed trajectory → checked physical motion.

Practice: twenty robotics Mathematics questions

1. Rotate point(1,0) by90° about the origin.
2. Translate point(1,2) by vector(4,3).
3. Rotate(1,0) by90° then translate by(4,3).
4. Explain why translating then rotating can give a different result.
5. For l₁=2,l₂=1,θ₁=30°,θ₂=60°, find end-effector x.
6. Find y for question5.
7. For link lengths2 and1, state reachable radial interval.
8. For l₁=l₂=1 and target(1,1), find cosθ₂.
9. For a two-link arm, what joint values make detJ=l₁l₂sinθ₂ zero?
10. vR=.8,vL=.4,b=.5. Find body speed.
11. Find angular speed.
12. Find ideal turning radius.
13. Both wheels=.5m/s for4s. Find distance.
14. Find length of path(0,0)→(3,4)→(6,4).
15. A circular robot radius.3m is point-planned. By how much should obstacles be inflated in the simple model?
16. For A* node with g=12,h=5, find f.
17. Evaluate q(s)=3s²−2s³ at s=.5.
18. From(1,1) to(4,5), find displacement distance.
19. Find bearing atan2(4,3) in degrees.
20. Why can coarse discrete collision checking miss an obstacle?

Worked answers

1. (0,1).
2. (5,5).
3. (4,4).
4. Because rigid-transform composition is order-sensitive; matrix multiplication is generally non-commutative.
5. About1.732.
6. 2.000.
7. 1≤r≤3.
8. 0.
9. θ₂=0 orπ modulo2π.
10. 0.6m/s.
11. 0.8rad/s.
12. 0.75m.
13. 2m.
14. 8 units.
15. 0.3m.
16. 17.
17. 0.5.
18. 5.
19. About53.13°.
20. Because a collision can lie between sampled points even when every sampled point is clear.

Sources and connected applications

For rigid-body transforms, forward kinematics, Jacobians and singularities, see Modern Robotics and MIT OpenCourseWare: Introduction to Robotics. For motion planning, see MIT Robotic Manipulation: Motion Planning. The robot dimensions and worked paths here are original teaching constructions.

Continue with Computer Graphics, Vectors, Matrices, Rendering and Interpolation; Timekeeping, Clocks, Calendars, Drift and Modular Arithmetic; and Supply Chains, Logistics, Network Flow, Routing and Warehouses. Return to the BTT Mathematics Hub.