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Tony's Intertwingled Memex β€’ Reference Manual

Mechanical Kinematics (Robotic Linkages) Solution

Formulas & Definitions

Problem Overview

We calculate the instantaneous tangent direction dydx\frac{dy}{dx}dxdy​ and normal cutting orientation for a CNC plasma cutter moving along an implicit elliptical track:

4x2+9y2=364x^2 + 9y^2 = 364x2+9y2=36
evaluated at the coordinate point (x,y)=(32,2)(x, y) = \left(\frac{3}{\sqrt{2}}, \sqrt{2}\right)(x,y)=(2​3​,2​).


Step-by-Step Mathematical Solution

Step 1: Formally Apply Implicit Differentiation

The governing equation is the Implicit Differentiation Chain Rule:

ddx[g(y)]=gβ€²(y)β‹…dydx\frac{d}{dx}[g(y)] = g'(y) \cdot \frac{dy}{dx}dxd​[g(y)]=gβ€²(y)β‹…dxdy​
(This rule is defined in Implicit Differentiation.)

Differentiate both sides of 4x2+9y2=364x^2 + 9y^2 = 364x2+9y2=36 with respect to xxx:

ddx[4x2]+ddx[9y2]=ddx[36]\frac{d}{dx}[4x^2] + \frac{d}{dx}[9y^2] = \frac{d}{dx}[36]dxd​[4x2]+dxd​[9y2]=dxd​[36]
8x+18ydydx=08x + 18y\frac{dy}{dx} = 08x+18ydxdy​=0

Step 2: Isolate the First Derivative dydx\frac{dy}{dx}dxdy​

Subtract 8x8x8x from both sides and divide by 18y18y18y:

18ydydx=βˆ’8xβ€…β€ŠβŸΉβ€…β€Šdydx=βˆ’8x18y=βˆ’4x9y18y\frac{dy}{dx} = -8x \implies \frac{dy}{dx} = -\frac{8x}{18y} = -\frac{4x}{9y}18ydxdy​=βˆ’8x⟹dxdy​=βˆ’18y8x​=βˆ’9y4x​

Step 3: Evaluate at the Operational Coordinates (32,2)\left(\frac{3}{\sqrt{2}}, \sqrt{2}\right)(2​3​,2​)

Substitute x=32x = \frac{3}{\sqrt{2}}x=2​3​ and y=2y = \sqrt{2}y=2​:

dydx∣(32,2)=βˆ’4(32)9(2)=βˆ’12292=βˆ’129β‹…2=βˆ’1218=βˆ’23\left.\frac{dy}{dx}\right|_{\left(\frac{3}{\sqrt{2}}, \sqrt{2}\right)} = -\frac{4\left(\frac{3}{\sqrt{2}}\right)}{9(\sqrt{2})} = -\frac{\frac{12}{\sqrt{2}}}{9\sqrt{2}} = -\frac{12}{9 \cdot 2} = -\frac{12}{18} = -\frac{2}{3}dxdy​​(2​3​,2​)​=βˆ’9(2​)4(2​3​)​=βˆ’92​2​12​​=βˆ’9β‹…212​=βˆ’1812​=βˆ’32​

Step 4: Determine the Perpendicular Normal Direction

The tangent slope of the track is mtangent=βˆ’23m_{\text{tangent}} = -\frac{2}{3}mtangent​=βˆ’32​. The normal line perpendicular to the track has slope:

mnormal=βˆ’1mtangent=βˆ’1βˆ’2/3=32=1.5m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{-2/3} = \frac{3}{2} = 1.5mnormal​=βˆ’mtangent​1​=βˆ’βˆ’2/31​=23​=1.5


Engineering Conclusion

At the position (32,2)\left(\frac{3}{\sqrt{2}}, \sqrt{2}\right)(2​3​,2​), the cutter head has a tangent track slope of βˆ’23-\frac{2}{3}βˆ’32​ (meaning for every 3Β meters3\text{ meters}3Β meters of horizontal travel in +x+x+x, the carriage moves 2Β meters2\text{ meters}2Β meters downward in βˆ’y-yβˆ’y). Industrial CNC controllers calculate the perpendicular normal slope mnormal=1.5m_{\text{normal}} = 1.5mnormal​=1.5 and continuously command the robot's wrist stepper motors to align the plasma torch perpendicular to the cut line. This orientation eliminates taper defects, guarantees clean edge bevels, and prevents expensive scrap in structural steel fabrication.