Industrial robot datasheets often list three linked metrics: payload (maximum end-effector load), speed (joint or tool velocity), and stiffness (resistance to deflection under load). Improving one tends to worsen at least one other because the same physical resources (actuator torque/power and structural mass/geometry) must satisfy all three.

flowchart TD
  A["Task requirement: payload, reach, cycle time, accuracy"] --> B["Statics check: torque >= load*lever"]
  B --> C["Actuator check: torque-speed curve + gear ratio"]
  C --> D["Structure check: deflection <= limit"]
  D --> E["Mass/inertia update (heavier links to increase stiffness)"]
  E --> F["Speed/acceleration update (J up => accel down)"]
  F --> G["Iterate until constraints satisfied"]
        

We will formalize each step with elementary models that you can interpret directly from a spec sheet.

Consider a single robot joint holding a payload at distance \( r \) from its axis. With payload mass \( m_p \) and link mass lumped at the tip \( m_\ell \), the gravitational load is \( (m_p+m_\ell)g \). The required holding torque is

\[ T_{\text{req}} = r (m_p+m_\ell) g . \]

Even before we talk about motion, a larger payload or longer reach increases torque linearly. If a robot is near its torque limit at full extension, it must move slower (or not at all) to avoid overload.

Define a payload margin as \( \mu_T = T_{\max}/T_{\text{req}} \). Safe operation requires \( \mu_T > 1 \). This margin shrinks when reach or payload rises.

A common actuator model (adequate at this conceptual stage) is the linear DC motor curve: with stall torque \( T_s \) and no-load speed \( \omega_0 \),

\[ T_m(\omega) = T_s\!\left(1 - \frac{\omega}{\omega_0}\right), \quad 0 \le \omega \le \omega_0 . \]

Now include a gearbox with ratio \( N \) (output torque multiplied by \(N\), output speed divided by \(N\)), and efficiency \( \eta \in (0,1] \). Joint torque and speed become

\[ T_j(\omega_j) = \eta N T_m(N\omega_j), \qquad \omega_j = \frac{\omega}{N}. \]

Substitute the motor curve:

\[ T_j(\omega_j) = \eta N T_s\!\left(1 - \frac{N\omega_j}{\omega_0}\right) = \eta N T_s - \eta \frac{N^2 T_s}{\omega_0}\omega_j . \]

Key consequence. Increasing \(N\) increases the zero-speed joint torque (payload capability) as \( \eta N T_s \), but decreases the maximum joint speed because the line hits zero torque at

\[ T_j(\omega_j)=0 \;\Rightarrow\; \omega_{j,\max} = \frac{\omega_0}{N}. \]

So the same actuator cannot simultaneously provide very high torque (for payload) and very high speed. Gearing shifts you along this trade-off curve.

A complementary view uses mechanical power: \( P_m(\omega)=T_m(\omega)\omega \). The maximum power for the linear curve occurs at mid-speed:

\[ P_m(\omega)=T_s\!\left(1-\frac{\omega}{\omega_0}\right)\omega, \quad \frac{dP_m}{d\omega} = 0 \Rightarrow \omega=\frac{\omega_0}{2}, \quad P_{\max}=\frac{T_s\omega_0}{4}. \]

Because gearing approximately conserves power (up to \(\eta\)), demanding high torque at the joint necessarily reduces speed.

To discuss stiffness without kinematics, approximate a robot link as a cantilever beam of length \( L \), modulus \( E \), and second moment of area \( I \). With an end force \( F \), the tip deflection is

\[ \delta = \frac{F L^3}{3 E I}. \]

The equivalent translational stiffness is \( k = F/\delta \), so

\[ k = \frac{3 E I}{L^3}. \]

Implications.

• For fixed geometry and material, longer reach \( L \) reduces stiffness as \( 1/L^3 \).
• Higher payload increases force \( F=(m_p+m_\ell)g \), increasing deflection linearly.

If accuracy requires \( \delta \le \delta_{\max} \), then

\[ \frac{(m_p+m_\ell)g L^3}{3 E I} \le \delta_{\max} \;\Rightarrow\; I \ge \frac{(m_p+m_\ell)g L^3}{3E\delta_{\max}} . \]

Meeting a stiffness/accuracy target at larger payload or reach demands larger \( I \), i.e., thicker/heavier links. That extra mass drives the next trade-off with speed.

To increase stiffness you typically add material (larger \(I\)), increasing mass. A rough lumped rotational inertia at the joint is \( J \approx m_{\text{eff}} L^2 \). Joint acceleration under torque \( T_j \) is

\[ \alpha = \frac{T_j}{J} \approx \frac{T_j}{m_{\text{eff}} L^2}. \]

If \(m_{\text{eff}}\) grows to meet stiffness, then \(\alpha\) drops, so the robot takes longer to reach a given speed (longer cycle time).

Stiffness also limits speed through vibration. The first-mode natural frequency is

\[ \omega_n = \sqrt{\frac{k}{m_{\text{eff}}}}. \]

Fast motions contain high-frequency components. To avoid exciting flexible modes, the motion/control bandwidth \( \omega_b \) should satisfy a separation condition such as

\[ \omega_b \le 0.2\,\omega_n . \]

Since \( \omega_n \) decreases when stiffness is low or mass is high, a compliant or heavy arm must be commanded more slowly. Your Linear Control background: this is the same idea as keeping controller bandwidth well below unmodeled flexible poles.

Imagine scaling a robot uniformly by factor \( s > 1 \): all lengths become \( sL \), and cross-sectional dimensions become \( s \) times larger.

(a) Stiffness scaling.

For similar shapes, the area moment scales as \( I \propto s^4 \), while length scales as \( L \propto s \). Using \( k = 3EI/L^3 \),

\[ k(s) \propto \frac{s^4}{(s)^3} = s. \]

Proof: replace \(I\) by \(s^4 I_0\) and \(L\) by \(s L_0\): \(k(s)=3E(s^4 I_0)/(s^3 L_0^3)=s(3E I_0/L_0^3)\). So stiffness rises only linearly with size.

(b) Mass and inertia scaling.

Volume scales as \( s^3 \), hence mass \( m(s) \propto s^3 \). Inertia about a joint scales roughly as

\[ J(s) \propto m(s)\,L(s)^2 \propto s^3 \cdot s^2 = s^5. \]

Thus, even though stiffness rises with size, inertia rises much faster. For similar actuator technology (torque density constant), available torque scales with motor volume: \( T_{\max}(s)\propto s^3 \). Therefore acceleration scales as

\[ \alpha(s) \propto \frac{T_{\max}(s)}{J(s)} \propto \frac{s^3}{s^5} = \frac{1}{s^2}. \]

Bigger robots can carry more (torque increases), but accelerate more slowly, which lowers practical speed for a given cycle time.

(c) Payload capacity scaling.

Payload force is limited by joint torque: \( F_{\max} \approx T_{\max}/L \). With the above scalings,

\[ F_{\max}(s) \propto \frac{s^3}{s} = s^2. \]

So payload grows quadratically with size, while acceleration drops quadratically: the essence of the trade-off.

We implement a simple model chaining: (i) motor torque–speed, (ii) gearing, (iii) static payload limit, (iv) beam deflection stiffness. These are not full robot simulations; they are spec-level calculations.

7.1 Python (NumPy/SciPy/Matplotlib)

import numpy as np

# Motor parameters (example)
T_s = 2.0       # stall torque [Nm]
w0 = 400.0      # no-load speed [rad/s]
eta = 0.85

# Geometry and material (example)
r = 0.6         # lever arm to payload [m]
L = 0.8         # link length [m]
E = 70e9        # Young modulus [Pa] (aluminum)
I = 2.5e-7      # second moment [m^4]
g = 9.81

def joint_curve(N, wj):
    # motor speed = N*wj
    wm = N*wj
    Tm = T_s*(1 - wm/w0)
    Tm = np.maximum(Tm, 0.0)
    return eta*N*Tm

def payload_limit(N, wj):
    Tj = joint_curve(N, wj)
    return Tj/(r*g)   # max (m_p+m_l) at that speed

def stiffness(E, I, L):
    return 3*E*I / (L**3)

def deflection(m_payload):
    F = m_payload*g
    return F*(L**3)/(3*E*I)

Ns = np.array([20, 40, 80, 120])
wj = np.linspace(0, 8, 200)

for N in Ns:
    mp_max = payload_limit(N, wj)
    print(f"N={N}, max payload at zero speed ~ {mp_max[0]:.2f} kg, max speed ~ {w0/N:.2f} rad/s")

k = stiffness(E, I, L)
print("Link stiffness k =", k, "N/m")
print("Deflection for 5 kg payload =", deflection(5.0), "m")
      

7.2 C++ (Eigen for vectors)

#include <iostream>
#include <Eigen/Dense>
using namespace Eigen;

int main() {
    double Ts = 2.0, w0 = 400.0, eta = 0.85;
    double r = 0.6, L = 0.8, E = 70e9, I = 2.5e-7, g = 9.81;

    auto jointTorque = [&](double N, double wj){
        double wm = N*wj;
        double Tm = Ts*(1.0 - wm/w0);
        if (Tm < 0) Tm = 0;
        return eta*N*Tm;
    };

    auto payloadLimit = [&](double N, double wj){
        return jointTorque(N, wj)/(r*g);
    };

    auto stiffness = [&](double E, double I, double L){
        return 3.0*E*I/(L*L*L);
    };

    Vector4d Ns(20,40,80,120);
    for (int i=0;i<Ns.size();++i){
        double N = Ns(i);
        double mp0 = payloadLimit(N, 0.0);
        double wjmax = w0/N;
        std::cout << "N=" << N << " mp_max(0)=" << mp0
                  << " kg, wj_max=" << wjmax << " rad/s\n";
    }

    double k = stiffness(E,I,L);
    std::cout << "k = " << k << " N/m\n";
    return 0;
}
      

7.3 Java (EJML or plain arrays)

public class TradeoffDemo {
    public static void main(String[] args) {
        double Ts = 2.0, w0 = 400.0, eta = 0.85;
        double r = 0.6, L = 0.8, E = 70e9, I = 2.5e-7, g = 9.81;

        java.util.function.BiFunction<Double, Double, Double> jointTorque =
            (N, wj) -> {
                double wm = N*wj;
                double Tm = Ts*(1.0 - wm/w0);
                if (Tm < 0) Tm = 0;
                return eta*N*Tm;
            };

        java.util.function.BiFunction<Double, Double, Double> payloadLimit =
            (N, wj) -> jointTorque.apply(N, wj)/(r*g);

        double[] Ns = {20,40,80,120};
        for (double N : Ns) {
            double mp0 = payloadLimit.apply(N, 0.0);
            double wjmax = w0/N;
            System.out.println("N=" + N + " mp_max(0)=" + mp0 +
                               " kg, wj_max=" + wjmax + " rad/s");
        }

        double k = 3.0*E*I/Math.pow(L,3);
        System.out.println("k = " + k + " N/m");
    }
}
      

7.4 MATLAB / Simulink (Robotics System Toolbox optional)

Ts = 2.0; w0 = 400.0; eta = 0.85;
r = 0.6; L = 0.8; E = 70e9; I = 2.5e-7; g = 9.81;

jointTorque = @(N,wj) eta*N*max(Ts*(1 - (N*wj)/w0), 0);
payloadLimit = @(N,wj) jointTorque(N,wj)/(r*g);

Ns = [20 40 80 120];
for N = Ns
    mp0 = payloadLimit(N,0);
    wjmax = w0/N;
    fprintf('N=%g mp_max(0)=%.2f kg, wj_max=%.2f rad/s\n', N, mp0, wjmax);
end

k = 3*E*I/L^3;
fprintf('k = %.3e N/m\n', k);

% In Simulink you can implement jointTorque and payloadLimit as MATLAB Function blocks,
% and sweep N to see payload-speed curves.
      

These scripts show how changing gear ratio moves you along a payload–speed curve, while stiffness depends strongly on reach and geometry.

Problem 1 (Gear Ratio for Payload):
A motor has stall torque \( T_s=1.5\,\mathrm{Nm} \), no-load speed \( \omega_0=300\,\mathrm{rad/s} \), and gearbox efficiency \( \eta=0.9 \). A link of lever arm \( r=0.5\,\mathrm{m} \) must hold a payload \( m_p=8\,\mathrm{kg} \) (ignore link mass). Find the minimum gear ratio \( N_{\min} \), and the resulting maximum joint speed.

Solution:

Required holding torque: \( T_{\text{req}} = r m_p g = 0.5 \cdot 8 \cdot 9.81 = 39.24\,\mathrm{Nm} \). Joint stall torque with gearing: \( T_{j,\text{stall}}=\eta N T_s \). Need \( \eta N T_s \ge T_{\text{req}} \):

\[ N_{\min} \ge \frac{T_{\text{req}}}{\eta T_s} = \frac{39.24}{0.9\cdot 1.5} \approx 29.07. \]

So \( N_{\min}\approx 30 \). Maximum joint speed: \( \omega_{j,\max}=\omega_0/N \approx 300/30=10\,\mathrm{rad/s} \).

Problem 2 (Stiffness Requirement):
A link of length \( L=1.0\,\mathrm{m} \) is modeled as a cantilever with modulus \( E=210\,\mathrm{GPa} \). Under payload force \( F=120\,\mathrm{N} \), the tip deflection must satisfy \( \delta \le 1.5\,\mathrm{mm} \). Find the minimum second moment \( I_{\min} \).

Solution:

Use \( \delta = F L^3/(3EI) \le \delta_{\max} \):

\[ I_{\min} \ge \frac{F L^3}{3E\delta_{\max}} = \frac{120\cdot 1^3}{3\cdot 210\times 10^9 \cdot 1.5\times 10^{-3}} \approx 1.27\times 10^{-7}\,\mathrm{m}^4. \]

If the current link has smaller \(I\), it must be thickened or changed to a stiffer section/material, increasing mass and reducing acceleration.

Problem 3 (Natural Frequency and Speed Limit):
A joint–link combination has equivalent stiffness \( k=2.0\times 10^5\,\mathrm{N/m} \) and effective moving mass \( m_{\text{eff}}=25\,\mathrm{kg} \). Compute \( \omega_n \), and a safe bandwidth limit \( \omega_b=0.2\,\omega_n \).

Solution:

\[ \omega_n=\sqrt{\frac{k}{m_{\text{eff}}}} =\sqrt{\frac{2.0\times 10^5}{25}} =\sqrt{8000}\approx 89.44\,\mathrm{rad/s}. \]

Safe bandwidth: \( \omega_b\approx 0.2\times 89.44 \approx 17.9\,\mathrm{rad/s} \). Motions faster than this are likely to excite flexural modes.

Problem 4 (Scaling Insight):
Show that if a robot is scaled uniformly by factor \(s\), with actuator torque density unchanged, payload capacity scales as \(s^2\) while acceleration scales as \(1/s^2\).

Solution:

Torque scales with motor volume \(\propto s^3\), reach scales as \(L\propto s\). Payload force limit \(F_{\max}\propto T_{\max}/L \propto s^3/s=s^2\). Inertia \(J\propto mL^2\propto s^3\cdot s^2=s^5\), so \(\alpha\propto T_{\max}/J\propto s^3/s^5=1/s^2\).

Payload, speed, and stiffness are coupled by simple physics. Payload needs torque \( T_{\text{req}}=r(m_p+m_\ell)g \); actuator torque–speed limits and gearing enforce a payload–speed trade-off: \( \omega_{j,\max}=\omega_0/N \). Stiffness follows structural scaling: \( k=3EI/L^3 \), so longer reach and higher payload increase deflection, forcing heavier links that raise inertia and lower acceleration. Finally, vibration sets a speed/bandwidth ceiling through \( \omega_n=\sqrt{k/m_{\text{eff}}} \).