Three common terms used in leadscrews are lead, # of starts, pitch. There definitions are usually as follows:

  • Lead usually refers to the amount of linear travel when the leadscrew turns one revolution.
  • # of Starts refers to the number of independent threads along the length of the leadscrew.
  • Pitch refers to the distance between threads. In leadscrew with one start, pitch = lead. With multiple starts pitch = lead / # starts.

In many leadscrew catalogues, the equation listed for translating a load's linear mass into angular inertia is:

$ J_{load} = m\times({\frac{1}{2\pi p}})^2$


$m$ is the mass of the load.

$p$ is the pitch in threads per inch.

My main issue here is that the units are threads per unit length instead of length per thread. The mix in nomenclature here is very confusing. In a leadscrew with one start, I can easily do $\frac{1}{lead}$ and plug that as $p$ into the equation. But what about when there are multiple starts? Do I still use $\frac{1}{lead}$, or $\frac{1}{\#ofstarts\times lead}$? What is the derivation of this equation?


1 Answer 1


Firstly, in reference to nomenclature: in ISO (metric) standards the pitch of a leadscrew is given in units of distance/revolution. However, Unified (inch series) standards, pitch is given in TPI (threads per inch). As you mentioned in your question, lead is related to pitch by $p = L/N$ where $p$ is the pitch, $L$ is the lead and $N$ is the number of starts.

Source: Juvinall & Marshek, Fundamentals of Machine Component Design, 5th Ed., pp. 412.

Here's a derivation of the equation. The equation relating linear load to torque for a lead screw is: $$ T = \frac{W d_m}{2}\frac{f\pi d_m + L \cos{\alpha_n}}{\pi d_m \cos{\alpha_n} - f L} $$ Where $T$ is the torque, $W$ is the linear load, $d_m$ is the mean diameter of thread contact, $f$ is the coefficient of friction, $L$ is the metric lead in distance/thread, and $\alpha_n$ is the thread angle measured in the normal plane.

Source: Juvinall & Marshek, Fundamentals of Machine Component Design, 5th Ed., pp. 418.

Let us examine the simplified case of a square thread ($\alpha_n = 0$). The above equation then simplifies to: $$T = \frac{W d_m}{2}\frac{f\pi d_m + L}{\pi d_m - f L} $$

Values of $f$ tend to be around 0.08 to 0.20 for steel against cast iron and bronze, and can be even lower for ballscrews (same source as above). Therefore let us simplify further by assuming $f = 0$. Then we get the following: $$ T = \frac{WL}{2\pi} $$

Now in this case our load $W= ma$ (mass times acceleration). Acceleration is related to angular acceleration by $a = \frac{L}{2\pi}\frac{d^2\theta}{dt^2}$. This gives: $$ T = \frac{mL^2}{(2\pi)^2}\frac{d^2\theta}{dt^2} $$ From the definition of rotational inertia ($T = J\frac{d^2\theta}{dt^2}$) we arrive at the final answer: $$ J = \left(\frac{L}{2\pi}\right)^2m$$ In other words, the torque you need to apply to move a linear load is independent of the number of starts of the leadscrew. For a single start, and using the Unified standard definition of pitch we get: $$ J = \frac{1}{(2\pi p)^2}m $$

The results of my derivation are supported by the catalogue I found that lists the calculation for the force-torque conversion for a lead screw (pp. 2, eq. 3 and the sample calculations on pp. 9) here. Note that they include the efficiency value $e$ which is used to approximate the friction in the system while simplifying the calculations. You should double check the equation given by whatever catalogue you are using.

Hope that helps.

  • $\begingroup$ I should remember adding references to my equations like you have heh. Now I can't even remember where I got the equation from my question $\endgroup$
    – DaveS
    Feb 2, 2017 at 13:05
  • $\begingroup$ @crxguy52 Good catch. I made an edit in response to your comment. $\endgroup$ Feb 6, 2019 at 22:38
  • $\begingroup$ @ConjuringFrictionForces is correct, but he missed a step when he converted form linear acceleration to rotation acceleration. Linear acceleration is m/s^2, angular acceleration is rad/sec^2. Therefore, $$a = \dfrac{L}{2\pi}\dfrac{d^2\theta}{dt^2}$$ Since pitch is distance/rev, this converts lead to distance/radian The equivalent inertia is then $$J = (\dfrac{L}{2\pi})^2m$$ The link in his post supports my case. $\endgroup$
    – crxguy52
    Feb 7, 2019 at 19:42

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