◀ Back to Note

Tony's Intertwingled Memex • Reference Manual

Structural Safe-Load Limits for Cargo Cranes (Mechanical Engineering) Solution

Formulas & Definitions

Problem Overview

We analyze a motorized shipyard gantry crane hoisting a cargo container along a vertical track. The winch drum is offset horizontally by a fixed distance x=24.0 mx = 24.0\text{ m}x=24.0 m. At a cable length z=30.0 mz = 30.0\text{ m}z=30.0 m, the winch reels cable inward at dzdt=−2.0 m/s\frac{dz}{dt} = -2.0\text{ m/s}dtdz​=−2.0 m/s. We calculate the vertical lift velocity dydt\frac{dy}{dt}dtdy​ and verify adherence to dynamic hoist load safety limits (vmax≤3.0 m/sv_{\text{max}} \le 3.0\text{ m/s}vmax​≤3.0 m/s).


Step-by-Step Mathematical Solution

Step 1: Geometric Setup & Differentiation

The cable geometry forms a right triangle where:

  • Horizontal drum offset x=24.0 mx = 24.0\text{ m}x=24.0 m is a static constant (dxdt=0\frac{dx}{dt} = 0dtdx​=0).
  • Vertical distance from drum level down to the cargo spreader is y(t)y(t)y(t) (dynamic variable).
  • Active cable length is the hypotenuse z(t)z(t)z(t) (dynamic variable).

The governing formula is the Pythagorean Theorem:

x2+y2=z2x^2 + y^2 = z^2x2+y2=z2
(This formula and its time derivative principles are defined in Related Rates.)

Substitute the static horizontal offset x=24.0x = 24.0x=24.0 before differentiating:

24.02+y2=z2  ⟹  576.0+y2=z224.0^2 + y^2 = z^2 \implies 576.0 + y^2 = z^224.02+y2=z2⟹576.0+y2=z2

Differentiating both sides implicitly with respect to time ttt using the Chain Rule:

0+2ydydt=2zdzdt  ⟹  ydydt=zdzdt0 + 2y \frac{dy}{dt} = 2z \frac{dz}{dt} \implies y \frac{dy}{dt} = z \frac{dz}{dt}0+2ydtdy​=2zdtdz​⟹ydtdy​=zdtdz​

Isolating the vertical velocity dydt\frac{dy}{dt}dtdy​:

dydt=zydzdt\frac{dy}{dt} = \frac{z}{y} \frac{dz}{dt}dtdy​=yz​dtdz​

Step 2: Snapshot Calculation of yyy and dydt\frac{dy}{dt}dtdy​

  1. Find the vertical position yyy when z=30.0 mz = 30.0\text{ m}z=30.0 m:
    576.0+y2=(30.0)2=900.0576.0 + y^2 = (30.0)^2 = 900.0576.0+y2=(30.0)2=900.0
    y2=900.0−576.0=324.0  ⟹  y=324.0=18.0 my^2 = 900.0 - 576.0 = 324.0 \implies y = \sqrt{324.0} = 18.0\text{ m}y2=900.0−576.0=324.0⟹y=324.0​=18.0 m
  2. Compute the instantaneous vertical velocity dydt\frac{dy}{dt}dtdy​:
    dydt=30.018.0(−2.0)=53(−2.0)=−103 m/s≈−3.33 m/s\frac{dy}{dt} = \frac{30.0}{18.0} (-2.0) = \frac{5}{3} (-2.0) = -\frac{10}{3}\text{ m/s} \approx -3.33\text{ m/s}dtdy​=18.030.0​(−2.0)=35​(−2.0)=−310​ m/s≈−3.33 m/s
    The negative sign indicates the vertical distance yyy from the gantry drum down to the container is decreasing, meaning the container is moving upward at a speed of 3.33 m/s3.33\text{ m/s}3.33 m/s.

Step 3: Structural Safety Compliance Evaluation

  • Regulatory Threshold: Maximum vertical hoist velocity vsafe≤3.0 m/sv_{\text{safe}} \le 3.0\text{ m/s}vsafe​≤3.0 m/s.
  • Calculated Vertical Velocity: ∣vvertical∣=3.33 m/s|v_{\text{vertical}}| = 3.33\text{ m/s}∣vvertical​∣=3.33 m/s.
  • Safety Verdict: NON-COMPLIANT / UNSAFE EXCURSION. The container's vertical ascent speed (3.33 m/s3.33\text{ m/s}3.33 m/s) exceeds the 3.0 m/s3.0\text{ m/s}3.0 m/s structural safety limit by 11%11\%11%, despite the winch reel speed being only 2.0 m/s2.0\text{ m/s}2.0 m/s.

Engineering Conclusion

In mechanical hoisting and crane automation, the winch spool tachometer measures only the line reel-in rate (dzdt\frac{dz}{dt}dtdz​), not the vertical cargo velocity (dydt\frac{dy}{dt}dtdy​).

Because dydt=zydzdt\frac{dy}{dt} = \frac{z}{y}\frac{dz}{dt}dtdy​=yz​dtdz​, the vertical velocity is amplified by the geometric factor zy=1cos⁡ϕ\frac{z}{y} = \frac{1}{\cos\phi}yz​=cosϕ1​ (where ϕ\phiϕ is the angle of the cable relative to the vertical track). As the container is hoisted higher, yyy decreases while the fixed horizontal offset x=24 mx = 24\text{ m}x=24 m remains constant. Consequently, the cable angle ϕ\phiϕ widens, and the ratio zy\frac{z}{y}yz​ increases rapidly:

  • When y=32 my = 32\text{ m}y=32 m (z=40 mz = 40\text{ m}z=40 m): zy=1.25  ⟹  ∣vy∣=2.5 m/s\frac{z}{y} = 1.25 \implies |v_y| = 2.5\text{ m/s}yz​=1.25⟹∣vy​∣=2.5 m/s (safe).
  • When y=18 my = 18\text{ m}y=18 m (z=30 mz = 30\text{ m}z=30 m): zy=1.67  ⟹  ∣vy∣=3.33 m/s\frac{z}{y} = 1.67 \implies |v_y| = 3.33\text{ m/s}yz​=1.67⟹∣vy​∣=3.33 m/s (exceeds 3.0 m/s3.0\text{ m/s}3.0 m/s limit).
  • When y=10 my = 10\text{ m}y=10 m (z=26 mz = 26\text{ m}z=26 m): zy=2.60  ⟹  ∣vy∣=5.20 m/s\frac{z}{y} = 2.60 \implies |v_y| = 5.20\text{ m/s}yz​=2.60⟹∣vy​∣=5.20 m/s (extreme hazard).

If crane control microcontrollers do not execute this related rates calculation in real time, a winch operator maintaining a constant motor spool speed of 2.0 m/s2.0\text{ m/s}2.0 m/s will unknowingly subject the crane structure and cargo rigging to severe shock accelerations and dynamic pendulum swinging. Microcontrollers must automatically derate and throttle down winch motor RPM as yyy decreases to enforce a strict cap on dydt\frac{dy}{dt}dtdy​.