<p><em><strong>This is a work in progress - I'm still plugging in the numbers</strong></em></p>
<hr />
<p>A couple of issues:</p>
<ul>
<li><strike>You appear to be mixing Force (effort) and Inertia and comparing them directly - this mistake seems to have come from the first link</strike>;</li>
<li>You used 8 mm <em>diameter</em>, instead of 4 mm <em>radius</em>, to calculate the torque</li>
<li>You haven't calculated the holding force, which would provide the holding torque required, which can be easily compared against the specification of the stepper motor;</li>
<li>I can't see where you obtained the value for axial force of 10 N, from the datasheet of the stepper motor.</li>
</ul>
<p>Regardless of these issues, tackling each question separately:</p>
<hr />
<blockquote>
<p>Selection stepper motor to lift given weight?</p>
</blockquote>
<h3>First equation</h3>
<p>Your first equation, for <em>effort</em>, is</p>
<p><span class="math-container">$$F_{effort} = S_f+\frac{Load}{2*\pi*\left(\frac{R}{P}\right)*S_e}$$</span></p>
<p>where,</p>
<ul>
<li><em>Starting Force</em>, from <a href="https://buildyourcnc.com/FAQ/879" rel="nofollow noreferrer">linked article</a>, worst case scenario, <span class="math-container">$S_f = 20$</span> N</li>
<li><em>Radius</em> <span class="math-container">$R=4$</span> mm<span class="math-container">$=0.004$</span> m</li>
<li><em>Pitch</em> <span class="math-container">$P=1.25$</span> mm<span class="math-container">$=0.0125$</span> m, and</li>
<li><em>Screw Efficiency</em>, from <a href="https://buildyourcnc.com/FAQ/879" rel="nofollow noreferrer">linked article</a>, worst case scenario 20 %, <span class="math-container">$S_e=20=0.2$</span></li>
</ul>
<p>So filling in the numbers, for 25 N:</p>
<p><span class="math-container">$$effort = 20+\frac{25}{2*\pi*\left(\frac{0.004}{0.00125}\right)*0.2}$$</span>
<span class="math-container">$$effort = 26.22$$</span></p>
<p>It is worth noting that this <em>running</em> torque only comes to 6.22 N, much smaller than the <em>worse case value</em> of the starting force, of 20 N:</p>
<p><span class="math-container">$$\frac{25}{2*\pi*\left(\frac{0.004}{0.00125}\right)*0.2}$$</span>
<span class="math-container">$$ = 6.22$$</span></p>
<p>and plugging in the numbers, for 50 N:</p>
<p><span class="math-container">$$effort = 20+\frac{50}{2*\pi*\left(\frac{0.004}{0.00125}\right)*0.2}$$</span>
<span class="math-container">$$effort = 32.44$$</span></p>
<p>Again, it is worth noting that this <em>running</em> torque only comes to 12.44 N, much smaller than the <em>worse case value</em> of the starting force, of 20 N:</p>
<p><span class="math-container">$$\frac{50}{2*\pi*\left(\frac{0.004}{0.00125}\right)*0.2}$$</span>
<span class="math-container">$$ = 12.44$$</span></p>
<p><strong>It should be noted</strong> that these values are for a force, in N, not torque, in N•m, and <em>I can't see on the <a href="http://bholanath.in/certificates-files/linear-actuator-TDS/BH42SH47-1504LA.pdf" rel="nofollow noreferrer">datasheet</a> were you obtained an Axial force of 10 N</em>. To convert these to a torque, you must multiply by the radius of the M8 x 1.25 spindle - M8 is 8 mm, therefore the radius is 4 mm, which is 0.004 m:</p>
<p><span class="math-container">$$T_{raise} = F_{effort} \times 0.004$$</span></p>
<p>Thus reducing the values for the Torque (derived from the effort) even further, by a couple of orders of magnitude, to 0.104 and 0.12976 N•m, for loads of 25 N and 50 N respectively.</p>
<p>So, the estimated value of <em>worst case</em> starting force would appear to be the dominant factor in selecting the stepper motor.</p>
<p>Regardless of all this, if, as you say, that the holding torque, <span class="math-container">$T_H$</span>, is higher than the running torque, <span class="math-container">$T_{raise}$</span>, then the holding torque would be the deciding factor in selecting a stepper and the running torque can be discounted.</p>
<p>It is unclear whether the equation that you found is the appropriate equation for calculating the holding torque. This equation appears to be for <em>moving</em> the load, whereas you need it for merely <em>holding</em> the load. For completeness both the holding and lowering torque values should be calculated.</p>
<h3>Second equation</h3>
<p>Your second equation apparently from <a href="https://en.wikipedia.org/wiki/Force#Rotations_and_torque" rel="nofollow noreferrer">Wikipedia - Force</a>, via <a href="https://www.daycounter.com/Calculators/Lead-Screw-Force-Torque-Calculator.phtml" rel="nofollow noreferrer">Lead Screw Torque and Force Calculator</a>,</p>
<p><span class="math-container">$$Torque_{raise} = F*D_m/2*\frac{L+\mu*\pi*D_m}{\pi*D_m-\mu*L}$$</span></p>
<p>where,</p>
<ul>
<li><em>Force</em>, <span class="math-container">$F$</span></li>
<li><em>Diameter</em>, <span class="math-container">$D_m$</span></li>
<li><em>Pitch distance of square thread</em>, <span class="math-container">$L=\frac{1}{D_{Thread}}$</span>, where <span class="math-container">$D_{Thread}$</span> is the thread density given in the table.</li>
<li><em>Coefficient of friction</em>, <span class="math-container">$\mu$</span></li>
</ul>
<p>Whilst I can not see that equation on Wikipedia's <a href="https://en.wikipedia.org/wiki/Force#Rotations_and_torque" rel="nofollow noreferrer">Force</a> page, I <em>did</em> manage to find the equation for raising, and the equation for lowering, on the Wikipedia page for <a href="https://en.wikipedia.org/wiki/Leadscrew#Mechanics" rel="nofollow noreferrer">Leadscrews - Mechanics</a><sup>1</sup>:</p>
<p><span class="math-container">$$T_R = \frac{Fd_m}{2}\left(\frac{l+\pi\mu{d_m}}{\pi{d_m}-\mu{l}}\right)$$</span></p>
<p>and</p>
<p><span class="math-container">$$T_L = \frac{Fd_m}{2}\left(\frac{\pi\mu{d_m}-l}{\pi{d_m}+\mu{l}}\right)$$</span></p>
<p>Plugging in the numbers for 50 N of load:</p>
<p><span class="math-container">$$T_R = \frac{50\times0.007375}{2}\left(\frac{0.00125+\pi\times 0.25\times0.007375}{\pi\times0.007375-0.25\times0.00125}\right)$$</span>
<span class="math-container">$$T_R = 0.1848 \times 0.308$$</span>
<span class="math-container">$$T_R = 0.0568$$</span></p>
<p>Plugging in the numbers for 50 N of load:</p>
<p><span class="math-container">$$T_L = \frac{50\times0.007375}{2}\left(\frac{\pi\times 0.25\times0.007375- 0.00125}{\pi\times0.007375+0.25\times0.00125}\right)$$</span>
<span class="math-container">$$T_L = 0.1848 \times 0.193$$</span>
<span class="math-container">$$T_L = 0.0357$$</span></p>
<p><strong>NOTE</strong>: As the link to the <a href="https://www.daycounter.com/Calculators/Lead-Screw-Force-Torque-Calculator.phtml" rel="nofollow noreferrer">calculator</a> states, there are two different torques - one to <em>raise</em>, <span class="math-container">$T_R$</span>, and one to <em>lower</em>, <span class="math-container">$T_L$</span>. As the lowering torque, <span class="math-container">$T_L$</span>, is <em>less than</em> the raising torque, <span class="math-container">$T_R$</span>, then it isn't <em>really necessary</em> to calculate it, or consider it when selecting the stepper, as the raising torque is the dominant factor.</p>
<p>Again, if, as you say, that the holding torque, <span class="math-container">$T_H$</span>, is higher than the running torque i.e. <span class="math-container">$T_R$</span>, then the holding torque would be the deciding factor in selecting a stepper and the running toque can be discounted.</p>
<hr />
<blockquote>
<p>Amount of load threaded rod can handle?</p>
</blockquote>
<p>This would appear to be much simpler to answer, as a value is given in the specifications for the rod, depending upon what you actually mean. It <em>could</em> be:</p>
<ul>
<li>The weight that the rod can support before slipping, or;</li>
<li>The weight that the rod can support before slipping before mechanical failure,</li>
</ul>
<p>I would imagine that, the specification given is the weight supported before the threads give way and break.</p>
<hr />
<h3>Footnotes</h3>
<p><sup>1</sup> These equations only account for the friction of the thread and the term for the collar appears to have been omitted.</p>
<p>From <a href="https://ncalculators.com/mechanical/torque-required-raise-load-calculator.htm" rel="nofollow noreferrer">Torque Required to Raise Load (TR) Calculation</a>, the <em>full</em> <a href="https://i.sstatic.net/3G23E.jpg" rel="nofollow noreferrer">equation</a>, for <em>both</em> thread and collar, is given by:</p>
<p><span class="math-container">$$T_R = \frac{Fd_m}{2}\left(\frac{l+\pi\mu{d_m}}{\pi{d_m}-\mu{l}}\right)+\left(\frac{F\mu_cd_c}{2}\right)$$</span></p>
<p>where,</p>
<ul>
<li><em>Torque</em> required to raise load, <span class="math-container">$T_R$</span></li>
<li><em>Load</em>, <span class="math-container">$F=50$</span> N</li>
<li><em>Mean diameter of square thread</em>, <span class="math-container">$d_m = X?$</span> m</li>
<li><em>Mean diameter of collar</em>, <span class="math-container">$d_c = X?$</span> m</li>
<li><em>Pitch distance of square thread</em>, <span class="math-container">$l = 0.00125$</span> m</li>
<li><em>Coefficient of friction for thread</em>, taken from the <strong><a href="https://ncalculators.com/mechanical/torque-required-raise-load-calculator.htm" rel="nofollow noreferrer">Solved Example</a></strong>, <span class="math-container">$\mu = 0.25$</span></li>
<li><em>Coefficient of friction of collar</em>, taken from the <strong><a href="https://ncalculators.com/mechanical/torque-required-raise-load-calculator.htm" rel="nofollow noreferrer">Solved Example</a></strong>, <span class="math-container">$\mu_c = 0.25$</span></li>
</ul>
<p>As the mean diameter, <span class="math-container">$d_m$</span> is not provided, it can be determined from the pitch, <span class="math-container">$l$</span>, and nominal diameter, <span class="math-container">$d_n = 8$</span> mm, by</p>
<p><span class="math-container">$$d_m = d_n-\frac{l}{2}$$</span>
<span class="math-container">$$d_m = 0.008 -\frac{0.00125}{2}$$</span>
<span class="math-container">$$d_m = 0.007375$$</span></p>
<p>The mean diameter of collar, <span class="math-container">$d_c$</span> is not given...</p>
<p>Plugging in the numbers for 50 N of load:</p>
<p><span class="math-container">$$T_R = \frac{50\times0.007375}{2}\left(\frac{0.00125+\pi\times 0.25\times0.007375}{\pi\times0.007375-0.25\times0.00125}\right)+\left(\frac{50\times0.25\times{d_c}}{2}\right)$$</span>
<span class="math-container">$$T_R = 0.1848 \times 0.308+\left(6.25\times{d_c}\right)$$</span>
<span class="math-container">$$T_R = 0.0568+\left(6.25\times{d_c}\right)$$</span></p>

CC BY-SA 4.0
2019-05-19T22:48:33.437Z
You made a mistake. Pitch is 0.00125
Vaidotas Rastenis
CC BY-SA 4.0
2019-05-19T23:09:02.317Z
Hi @VaidotasRastenis and welcome to SE.3DP. Ah yes, so I did...many thanks... I'll convert your answer to a comment to my answer. Thank you again. :-) Also the radius of the spindle should be 0.004, not 0.04. Fortunately, that means the answers to the first equation remain the same. The result to second equation changes though.
Greenonline
CC BY-SA 4.0
2020-08-03T10:15:17.590Z
@Trish - ah yes, wow, I'd forgotten all about this. Hmm, I wish I'd never attempted to answer it... it's a nightmare of a question... lol. Would you like to finish it off for me? :-)
Greenonline
CC BY-SA 4.0
2020-06-08T19:00:05.840Z
oldy but goldie - still WIP?
Trish
CC BY-SA 4.0
2024-03-12T20:09:07.767Z
@Trish - Apparently,  [this comment](https://engineering.stackexchange.com/questions/28460/mean-diameter-of-the-collar-when-calculating-the-raising-torque#comment51781_28460) on [my question over on Engineering.SE](https://engineering.stackexchange.com/q/28460/5899) sort of implies that either: the question can not be answered (which I'm not sure is correct), or; I was going about it the wrong way and using the wrong equation. Maybe it should be closed as a duplicate then, if your linked question is the same thing.
Greenonline
CC BY-SA 4.0
2019-04-30T17:50:00.567Z
2019-05-20T11:48:40.570Z
2020-06-18T08:29:41.667Z
Greenonline
<p>I am trying to figure out some calculations regarding stepper motor and lead screw.</p>

<p>These are the things need answer:</p>

<ol>
<li>Selection stepper motor to lift given weight?</li>
<li>Amount of load threaded rod can handle?</li>
</ol>

<p>I found a this topic <a href="https://buildyourcnc.com/FAQ/879" rel="nofollow noreferrer">how much weight can my stepping motor lift?</a></p>

<p><a href="https://i.sstatic.net/8qpvW.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/8qpvW.png" alt="Equation"></a></p>

<p>where, </p>

<ul>
<li>p = pitch of the screw </li>
<li>Se = screw efficiency = Standard lead screw will be between 20% (.2) and 40% (.4) </li>
<li>Sf = static force. This is the force that is needed to start the movement. The number may be eliminated, but it is good to use a number in the 5 N to 20 N range. </li>
<li>Load = the expected load that the effort will need to carry (i.e., the router and the included axis assembly that the motor will need to lift) </li>
<li>R = radius of the lead screw</li>
</ul>

<p>Based on this equation with following data</p>

<ul>
<li><p>Load: 5&nbsp;Kg = ~25&nbsp;N</p></li>
<li><p>Thread rod specs;</p>

<ul>
<li>Pitch : 1.25 mm = 0.0492126</li>
<li>Dia : 8 mm = 0.314961 inch</li>
</ul></li>
<li><p>Motor :</p>

<ul>
<li>Nema 17 </li>
<li><a href="https://steppermotors.bholanath.in/stepper-motors/captive-lead-screw-linear-actuator-stepper-motors-length-250mm-300mm-450mm/5-5-kg-cm-captive-linear-actuator-stepper-motor-1-5-amp-motor-pitch-m8x1-25-with-250-mm-lead-screw.html" rel="nofollow noreferrer">Datasheet</a></li>
</ul></li>
</ul>

<p>Calculated <strong>Effort = 6.22 N = 1.4 lbf = 22.4 ounce</strong></p>

<p>Radius = 8&nbsp;mm = 0.315 inch (using <a href="https://www.newmantools.com/tech/pitchconversions.htm" rel="nofollow noreferrer">Pitch conversion table</a>)</p>

<p>So torque = 22.4 * 0.315 = 7.056 oz-in = 0.0498&nbsp;Nm </p>

<p>As per data sheet Motor can provide an Axial force of 10&nbsp;N. I'm using two of such motors, so this motor is enough</p>

<p>However, based on this calculator, <a href="https://www.daycounter.com/Calculators/Lead-Screw-Force-Torque-Calculator.phtml" rel="nofollow noreferrer">Lead Screw Torque and Force Calculator</a>, with following information:</p>

<blockquote>
  <p><a href="https://i.sstatic.net/vXF5N.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/vXF5N.png" alt="Lead Screw Torque and Force Calculator"></a></p>
</blockquote>

<p>Torque required to raise in 0.0252&nbsp;Nm.</p>

<ul>
<li>Which should be used for calculations?</li>
<li>How to find the running torque of stepper motor? </li>
<li>I know it will be less that Holding torque, but what percentage?</li>
</ul>

<p>Nema 17 that I've given Holding torque is 5.5&nbsp;Kgcm or 0.55&nbsp;Nm. My second calculation gives the calculated torque to lift is 0.0252&nbsp;Nm.</p>

<p>Also from this <a href="https://www.engineeringtoolbox.com/loads-hanging-rods-d_1341.html" rel="nofollow noreferrer">Threaded rod load</a> table, load capacity of 8&nbsp;mm rod is somewhat around 270&nbsp;Kg.</p>

<p>Looking for further inputs.</p>

CC BY-SA 4.0
2018-12-18T15:23:04.327Z
5 kg is about 50 N, maybe change `Load: 5 Kg = ~25 N` to `Load: 5 Kg  ~ 50 N => 25 N per stepper`
0scar
CC BY-SA 4.0
2018-12-18T20:23:06.973Z
Possible duplicate of [Ball screw: Maintaining Z axis position when motor is OFF](https://3dprinting.stackexchange.com/questions/6949/ball-screw-maintaining-z-axis-position-when-motor-is-off) - Same problem, different values
Trish
CC BY-SA 4.0
2024-03-12T20:18:05.147Z
I've closed as a duplicate as the previous 4 votes had aged away, but this question doesn't seem to garner any answers, and I don't know how to finish my answer off. If you believe it isn't a dupe then feel free to vote to reopen, explaining why it isn't a duplicate...
Greenonline
<z-axis><stepper><lead-screw>
Z axis: Stepper motor and lead screw torque calculation
Athul
# Z axis: Stepper motor and lead screw torque calculation

(Asked by: username_0 on 2018-12-18)

I am trying to figure out some calculations regarding stepper motor and lead screw.

These are the things need answer:

1. Selection stepper motor to lift given weight?
2. Amount of load threaded rod can handle?

I found a this topic [how much weight can my stepping motor lift?](https://buildyourcnc.com/FAQ/879)

[![Equation](https://i.sstatic.net/8qpvW.png)](https://i.sstatic.net/8qpvW.png)

where,

* p = pitch of the screw
* Se = screw efficiency = Standard lead screw will be between 20% (.2) and 40% (.4)
* Sf = static force. This is the force that is needed to start the movement. The number may be eliminated, but it is good to use a number in the 5 N to 20 N range.
* Load = the expected load that the effort will need to carry (i.e., the router and the included axis assembly that the motor will need to lift)
* R = radius of the lead screw

Based on this equation with following data

* Load: 5 Kg = ~25 N
* Thread rod specs;

  + Pitch : 1.25 mm = 0.0492126
  + Dia : 8 mm = 0.314961 inch
* Motor :

  + Nema 17
  + [Datasheet](https://steppermotors.bholanath.in/stepper-motors/captive-lead-screw-linear-actuator-stepper-motors-length-250mm-300mm-450mm/5-5-kg-cm-captive-linear-actuator-stepper-motor-1-5-amp-motor-pitch-m8x1-25-with-250-mm-lead-screw.html)

Calculated **Effort = 6.22 N = 1.4 lbf = 22.4 ounce**

Radius = 8 mm = 0.315 inch (using [Pitch conversion table](https://www.newmantools.com/tech/pitchconversions.htm))

So torque = 22.4 \* 0.315 = 7.056 oz-in = 0.0498 Nm

As per data sheet Motor can provide an Axial force of 10 N. I'm using two of such motors, so this motor is enough

However, based on this calculator, [Lead Screw Torque and Force Calculator](https://www.daycounter.com/Calculators/Lead-Screw-Force-Torque-Calculator.phtml), with following information:

> [![Lead Screw Torque and Force Calculator](https://i.sstatic.net/vXF5N.png)](https://i.sstatic.net/vXF5N.png)

Torque required to raise in 0.0252 Nm.

* Which should be used for calculations?
* How to find the running torque of stepper motor?
* I know it will be less that Holding torque, but what percentage?

Nema 17 that I've given Holding torque is 5.5 Kgcm or 0.55 Nm. My second calculation gives the calculated torque to lift is 0.0252 Nm.

Also from this [Threaded rod load](https://www.engineeringtoolbox.com/loads-hanging-rods-d_1341.html) table, load capacity of 8 mm rod is somewhat around 270 Kg.

Looking for further inputs.

**Tags:** z-axis, stepper, lead-screw

> **Comments:**
> **username_1**: 5 kg is about 50 N, maybe change `Load: 5 Kg = ~25 N` to `Load: 5 Kg  ~ 50 N => 25 N per stepper`  
> **username_2**: Possible duplicate of [Ball screw: Maintaining Z axis position when motor is OFF](https://3dprinting.stackexchange.com/questions/6949/ball-screw-maintaining-z-axis-position-when-motor-is-off) - Same problem, different values  
> **username_3**: I've closed as a duplicate as the previous 4 votes had aged away, but this question doesn't seem to garner any answers, and I don't know how to finish my answer off. If you believe it isn't a dupe then feel free to vote to reopen, explaining why it isn't a duplicate...  

---

## 1 answers

### Answer by username_3 on 2019-04-30. Score: 1

***This is a work in progress - I'm still plugging in the numbers***

---

A couple of issues:

* You appear to be mixing Force (effort) and Inertia and comparing them directly - this mistake seems to have come from the first link;
* You used 8 mm *diameter*, instead of 4 mm *radius*, to calculate the torque
* You haven't calculated the holding force, which would provide the holding torque required, which can be easily compared against the specification of the stepper motor;
* I can't see where you obtained the value for axial force of 10 N, from the datasheet of the stepper motor.

Regardless of these issues, tackling each question separately:

---

> Selection stepper motor to lift given weight?

### First equation

Your first equation, for *effort*, is

$$F\_{effort} = S\_f+\frac{Load}{2\*\pi\*\left(\frac{R}{P}\right)\*S\_e}$$

where,

* *Starting Force*, from [linked article](https://buildyourcnc.com/FAQ/879), worst case scenario, $S\_f = 20$ N
* *Radius* $R=4$ mm$=0.004$ m
* *Pitch* $P=1.25$ mm$=0.0125$ m, and
* *Screw Efficiency*, from [linked article](https://buildyourcnc.com/FAQ/879), worst case scenario 20 %, $S\_e=20=0.2$

So filling in the numbers, for 25 N:

$$effort = 20+\frac{25}{2\*\pi\*\left(\frac{0.004}{0.00125}\right)\*0.2}$$
$$effort = 26.22$$

It is worth noting that this *running* torque only comes to 6.22 N, much smaller than the *worse case value* of the starting force, of 20 N:

$$\frac{25}{2\*\pi\*\left(\frac{0.004}{0.00125}\right)\*0.2}$$
$$ = 6.22$$

and plugging in the numbers, for 50 N:

$$effort = 20+\frac{50}{2\*\pi\*\left(\frac{0.004}{0.00125}\right)\*0.2}$$
$$effort = 32.44$$

Again, it is worth noting that this *running* torque only comes to 12.44 N, much smaller than the *worse case value* of the starting force, of 20 N:

$$\frac{50}{2\*\pi\*\left(\frac{0.004}{0.00125}\right)\*0.2}$$
$$ = 12.44$$

**It should be noted** that these values are for a force, in N, not torque, in N•m, and *I can't see on the [datasheet](http://bholanath.in/certificates-files/linear-actuator-TDS/BH42SH47-1504LA.pdf) were you obtained an Axial force of 10 N*. To convert these to a torque, you must multiply by the radius of the M8 x 1.25 spindle - M8 is 8 mm, therefore the radius is 4 mm, which is 0.004 m:

$$T\_{raise} = F\_{effort} \times 0.004$$

Thus reducing the values for the Torque (derived from the effort) even further, by a couple of orders of magnitude, to 0.104 and 0.12976 N•m, for loads of 25 N and 50 N respectively.

So, the estimated value of *worst case* starting force would appear to be the dominant factor in selecting the stepper motor.

Regardless of all this, if, as you say, that the holding torque, $T\_H$, is higher than the running torque, $T\_{raise}$, then the holding torque would be the deciding factor in selecting a stepper and the running torque can be discounted.

It is unclear whether the equation that you found is the appropriate equation for calculating the holding torque. This equation appears to be for *moving* the load, whereas you need it for merely *holding* the load. For completeness both the holding and lowering torque values should be calculated.

### Second equation

Your second equation apparently from [Wikipedia - Force](https://en.wikipedia.org/wiki/Force#Rotations_and_torque), via [Lead Screw Torque and Force Calculator](https://www.daycounter.com/Calculators/Lead-Screw-Force-Torque-Calculator.phtml),

$$Torque\_{raise} = F\*D\_m/2\*\frac{L+\mu\*\pi\*D\_m}{\pi\*D\_m-\mu\*L}$$

where,

* *Force*, $F$
* *Diameter*, $D\_m$
* *Pitch distance of square thread*, $L=\frac{1}{D\_{Thread}}$, where $D\_{Thread}$ is the thread density given in the table.
* *Coefficient of friction*, $\mu$

Whilst I can not see that equation on Wikipedia's [Force](https://en.wikipedia.org/wiki/Force#Rotations_and_torque) page, I *did* manage to find the equation for raising, and the equation for lowering, on the Wikipedia page for [Leadscrews - Mechanics](https://en.wikipedia.org/wiki/Leadscrew#Mechanics)1:

$$T\_R = \frac{Fd\_m}{2}\left(\frac{l+\pi\mu{d\_m}}{\pi{d\_m}-\mu{l}}\right)$$

and

$$T\_L = \frac{Fd\_m}{2}\left(\frac{\pi\mu{d\_m}-l}{\pi{d\_m}+\mu{l}}\right)$$

Plugging in the numbers for 50 N of load:

$$T\_R = \frac{50\times0.007375}{2}\left(\frac{0.00125+\pi\times 0.25\times0.007375}{\pi\times0.007375-0.25\times0.00125}\right)$$
$$T\_R = 0.1848 \times 0.308$$
$$T\_R = 0.0568$$

Plugging in the numbers for 50 N of load:

$$T\_L = \frac{50\times0.007375}{2}\left(\frac{\pi\times 0.25\times0.007375- 0.00125}{\pi\times0.007375+0.25\times0.00125}\right)$$
$$T\_L = 0.1848 \times 0.193$$
$$T\_L = 0.0357$$

**NOTE**: As the link to the [calculator](https://www.daycounter.com/Calculators/Lead-Screw-Force-Torque-Calculator.phtml) states, there are two different torques - one to *raise*, $T\_R$, and one to *lower*, $T\_L$. As the lowering torque, $T\_L$, is *less than* the raising torque, $T\_R$, then it isn't *really necessary* to calculate it, or consider it when selecting the stepper, as the raising torque is the dominant factor.

Again, if, as you say, that the holding torque, $T\_H$, is higher than the running torque i.e. $T\_R$, then the holding torque would be the deciding factor in selecting a stepper and the running toque can be discounted.

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> Amount of load threaded rod can handle?

This would appear to be much simpler to answer, as a value is given in the specifications for the rod, depending upon what you actually mean. It *could* be:

* The weight that the rod can support before slipping, or;
* The weight that the rod can support before slipping before mechanical failure,

I would imagine that, the specification given is the weight supported before the threads give way and break.

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### Footnotes

1 These equations only account for the friction of the thread and the term for the collar appears to have been omitted.

From [Torque Required to Raise Load (TR) Calculation](https://ncalculators.com/mechanical/torque-required-raise-load-calculator.htm), the *full* [equation](https://i.sstatic.net/3G23E.jpg), for *both* thread and collar, is given by:

$$T\_R = \frac{Fd\_m}{2}\left(\frac{l+\pi\mu{d\_m}}{\pi{d\_m}-\mu{l}}\right)+\left(\frac{F\mu\_cd\_c}{2}\right)$$

where,

* *Torque* required to raise load, $T\_R$
* *Load*, $F=50$ N
* *Mean diameter of square thread*, $d\_m = X?$ m
* *Mean diameter of collar*, $d\_c = X?$ m
* *Pitch distance of square thread*, $l = 0.00125$ m
* *Coefficient of friction for thread*, taken from the **[Solved Example](https://ncalculators.com/mechanical/torque-required-raise-load-calculator.htm)**, $\mu = 0.25$
* *Coefficient of friction of collar*, taken from the **[Solved Example](https://ncalculators.com/mechanical/torque-required-raise-load-calculator.htm)**, $\mu\_c = 0.25$

As the mean diameter, $d\_m$ is not provided, it can be determined from the pitch, $l$, and nominal diameter, $d\_n = 8$ mm, by

$$d\_m = d\_n-\frac{l}{2}$$
$$d\_m = 0.008 -\frac{0.00125}{2}$$
$$d\_m = 0.007375$$

The mean diameter of collar, $d\_c$ is not given...

Plugging in the numbers for 50 N of load:

$$T\_R = \frac{50\times0.007375}{2}\left(\frac{0.00125+\pi\times 0.25\times0.007375}{\pi\times0.007375-0.25\times0.00125}\right)+\left(\frac{50\times0.25\times{d\_c}}{2}\right)$$
$$T\_R = 0.1848 \times 0.308+\left(6.25\times{d\_c}\right)$$
$$T\_R = 0.0568+\left(6.25\times{d\_c}\right)$$

> **Comments:**
> **username_4**: You made a mistake. Pitch is 0.00125  
> **username_3**: Hi @username_4 and welcome to SE.3DP. Ah yes, so I did...many thanks... I'll convert your answer to a comment to my answer. Thank you again. :-) Also the radius of the spindle should be 0.004, not 0.04. Fortunately, that means the answers to the first equation remain the same. The result to second equation changes though.  
> **username_2**: oldy but goldie - still WIP?  
> **username_3**: @username_2 - ah yes, wow, I'd forgotten all about this. Hmm, I wish I'd never attempted to answer it... it's a nightmare of a question... lol. Would you like to finish it off for me? :-)  
> **username_3**: @username_2 - Apparently,  [this comment](https://engineering.stackexchange.com/questions/28460/mean-diameter-of-the-collar-when-calculating-the-raising-torque#comment51781_28460) on [my question over on Engineering.SE](https://engineering.stackexchange.com/q/28460/5899) sort of implies that either: the question can not be answered (which I'm not sure is correct), or; I was going about it the wrong way and using the wrong equation. Maybe it should be closed as a duplicate then, if your linked question is the same thing.  

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