Mechanical Engineering

First determine what range of weight you are moving, speed, so you can determine the thread pitch and type of motor used

let's say it is 1 Kg . what is the value of the torque we need?

I answered this question in detail just a few days ago.

Better vendors provide required torque calculators for exerting any reasonable ultimate force with any prospective leadscrew configuration. In other words, these calculators will account for the characteristic resistance and mechanical advantages imposed by different screw types, diameters, pitches, and sometimes even finishes.

In my previous answer, I attempted to upload an Excel template for making these calculations, but GC rules forbid accepting any permutation of my file (including zipped instances). I would encourage you to follow at least the idea however, because you will inevitably compare different prospective combinations of motors, gearing, and leadscrews to identify an ideal combination for your purposes. I prefer to integrate the comparisons in a single spreadsheet, so that you have a permanent record of the determining parameters of your ultimate decision. The formulas for making the calculations are readily available.

My own spreadsheets have well-refined structures for calculating the usual plurality of instrumental scenarios. Because GC rules forbid uploading them , I did a little reverse digging and turned up a few of my old links. You will need to combine some of these formulae (whichever your calculations require) in your own spreadsheet calculator.

This calculator determines the maximum power and speed of stepper motors:

https://www.allaboutcircuits.com/tools/stepper-motor-calculator/

Here is a spreadsheet download that may help you further. I much prefer my own. But if I recall correctly, this example will make a good building block:

https://triquetra-cnc.com/stepper-calibration-download/

Here are further means to make further calculations which you may end up leaning upon for further related purposes (good at least to bookmark):

https://www.globalspec.com/pfdetail/motors/stepper-motor-calculations

Here also is a search turning up a multitude of calculators which determine the force required to lift a given weight (or exert a given force) with various prospective leadscrews:

https://duckduckgo.com/?t=ffab&q=how+to+calculate+the+force+required+to+lift+weight+with+a+leadscrew&atb=v166-1&ia=web

Your basic problem or calculation therefore reduces to this:

You have some ultimate force you mean to exert, through whatever leadscrew and gearing combinations it would be advisable to consider deploying.

Each prospective combination of leadscrew and gearing thus determines the minimal motor force your mechanical capacity will require.

Thus, you first determine the ultimate force which is required to perform your intended movement. Then you determine the force required to turn an ideal leadscrew for exerting this ultimate required force. Finally then, you determine how much force will exert that much turning force on your leadscrew, acting through a prospective transmission (if any). So, you may either drive your leadscrew directly from a motor, or succeed in utilizing even smaller motors, to drive your leadscrew by way of a transmission.

If for example, you need one Newton of force to turn your leadscrew with such force as will exert the ultimate force of your mechanism, then approximately 1/50 of a Newton will be required to turn your leadscrew through a 50:1 gear reduction box (transmission). So, the minimum force you may require from your motor (driving that gearbox) will be at least 1/50 of a Newton (because of further friction and drag introduced by the gearbox, etc.).

Thus for the usual reasons, it will generally be advisable to provide your mechanical capacity more power than this. In other words, expect it to be more advisable to choose a motor which will deliver perhaps at least 150-200% of the minimal force required to exert your ultimate force through the leadscrew and gearing.

To account for the additional resistance introduced by the gearbox then, your minimum required motor power calculation should be increased by some percentage.

It is also wise to further increase the calculation however, because surplus power will further ensure that your intended movements prevail against exceptional or potentially extreme circumstances. Thus your spreadsheet may multiply the calculated minimal power requirement by perhaps 2 (or 200%). Because of the ranges of power provided by available motors, a best commercial match (meeting your minimal required power calculation) will often inherit such an advantage.

But it is regularly better to utilize motors at some smaller fraction of their maximum deliverable power. You realize for example, the escalating encumbrances we suffer in running more difficult wood through a marginally powered 10-inch table saw. As the engine bogs, feed rate must be reduced... and you don't want to trap yourself into needing altogether to dynamically detect the exactitudes of such undesirable aberrations, to calculate the required adjustments in feed, and to make the required provisions in your software. Ten bucks of bigger motor not only avoids all the additional and far more technical software code and hardware required; it avoids the undesirable condition altogether.

Stepper motors are likewise subject to characteristic power bands which mean they have much less power at higher RPMs. So, it is really the minimum power required at the RPMs which are required to deliver that power, that we are concerned with. To avoid potential issues, a sufficient rule of thumb is simply to step your power well upward of your minimal required power calculations. Nonetheless, good minimal required power calculations are indispensable to determining the required ballpark.

Obviously, finer screw pitches (and even finer screw diameters) more efficiently exert the force of your motor; and lower gearing can greatly increase the potential force exerted — while both finer screw pitches and lower gearing ratios have calculable expenses in potential speed.

Thus, having determined motor, gearing, and leadscrew combinations which will exert at least the ultimate force your mechanism requires; then finally, with the first group of formulas presented above, we may now calculate the upper reaches of advisable speeds for rotating each decided motor, and determine how fast your resultant mechanical capacities will move objects or apparatus with the force you are now capable of exerting.

To be useful, these instructions must be at least this clear and detailed. If you carefully follow them, I expect you will finish the job with less work or difficulty than you might first anticipate. But I'm sure you already appreciate then, how appropriate and conducive spreadsheets are, not only for making the required diversity of calculations, but also for reference and record-keeping purposes.

In reviewing both permutations of your question, I believe I should make one more point:

Your initial question states only that you want to "move" a mass, m. I am not sure you are aware then of the differences between moving m horizontally for example (either with or without further resistance,) or lifting m.

You later rephrase your initial question, merely to specify the weight you want to move. No one can answer your question still however, because to calculate the motor force required to turn the leadscrew, you must also provide at least the diameter and pitch of the leadscrew you may or may not have selected.

CALCULATING THE FORCE TO TURN A PRE-DETERMINED LEADSCREW

You may be motorizing a pre-existent mechanism or leadscrew. Or, you may already have well determined an ideal leadscrew.

For a predetermined leadscrew, you would simply use any of the leadscrew calculators in the links I provided, to calculate required torque. This group of calculators determines the force required to turn your leadscrew diameter and pitch, so that it will in turn exert the ultimate force, uf, you mean to exert.

OTHERWISE DETERMINING AN IDEAL LEADSCREW

If on the further hand, the leadscrew pitch, diameter, and finish are not yet determined, you would perform the same calculations for every prospective leadscrew, until an ideal leadscrew and motor combination emerges from the process.

All in all, smaller leadscrew diameters and more moderate leadscrew pitches provide greater mechanical advantages to generate the required force. Thus, smaller diameters and more moderate pitches require less motor power.

But the force you must generate thru the length of the leadscrew also determines a minimum permissible leadscrew diameter; and you must first identify this minimum permissible diameter, because it will comprise the bottom (or smallest) end of the prospective leadscrews you may utilize to exert your force. Therefore, the length of the leadscrew you require also affects this determination. The greater its length, the heavier the leadscrew diameter must be.

Thus, you would choose your minimum conceivable leadscrew diameter from tables, ranking the capacities of available leadscrews to perform the work you require; and so, the structure of any such truly comprehensive tables will distinguish ranges of leadscrew lengths, capable of exerting your force uf, with each prospective diameter and pitch.

Vendors and makers ought to make themselves sources of this determinate information. But lacking sufficiently good resources, you may have to model your implementation after similar, proven models. If you see a similar mechanism move as much mass as you intend to move your mass, the observed equivalences may infer the propriety of that leadscrew diameter and pitch, as long as all other factors are sufficiently accounted for. Excellent common sense, geometry, and mathematics serve to make more intricate comparisons.

Generally then, as smaller diameters impose less resistance to turning, but also because smaller diameters lend to the more moderate pitches which provide greater mechanical advantages to exert your ultimate force, the smallest feasible diameters will tend to be the best candidates for minimal motor power, unless a particular diameter may not offer the finest pitches which other diameters may make available.

THE DIFFERENCE BETWEEN "MOVING" m AND LIFTING m

Equally critical to answering your question however, is the fact that there is a difference between "moving" and lifting which we are required to observe. This distinguishing consideration is that a force greater than m is required to lift m; and that any force at all (thus prospectively, a tiny fraction of m) may succeed in "moving" m otherwise, depending upon further prospective resistance.

Thus to lift m, the minimum ultimate force you need to exert must be greater than m, because a vertical upward force of m will only sustain its elevation. Likewise, a force less than m will instead allow m to accelerate downwardly.

A very small force on the other hand, may succeed in moving m horizontally for example — particularly if further resistance is negligible. The ratio of the force to the mass m, multiplied by the acceleration of gravity, determines how long it will take your force to accelerate m to whatever velocity is required for the movements of your mechanism.

DETERMINING THE FORCE TO MOVE m HORIZONTALLY

Supposing that m has no further resistance (such as is exerted by gravity), then you may move m with a much smaller force than m; and how much you may accelerate m with that prospectively much smaller force is predicated by the acceleration of gravity. In other words, with the aforesaid lack of any further resistance, a force of m exerted against m will horizontally accelerate m 32 feet per second, per second. Therefore, a force of m/2 will accelerate m 16 fps per second; a force of m/32 will accelerate m 1 fps per second; and so forth.

The formula for determining an ultimate force uf to horizontally accelerate m to v fps in one second would therefore be:

uf = m × ( v ÷ 32)

Thus if m is your 1 Kg, and you want to accelerate m to 10 fps in 1 second (or 5 fps in half a second and so forth), then your calculation is:

uf = 1 Kg × ( 10 ÷ 32 ) = 0.3125 Kg

So, to determine the horizontal force required to move m against no other resistance, you first estimate the uppermost velocity of the movement you wish to accomplish.

Let's say you want to move your Kg 4 inches, and that you visualize this maneuver transpiring in 1 second.

Recognizing that the entire movement may well transpire under acceleration then, the rate of travel at the midpoint of the 4 inches must be your 4 inches per second required overall rate of travel, because the velocity at the moment you initiate the yet-undetermined rate of acceleration is 0 ips. Thus, given the deficient average velocity over the first half of the intended maneuver, we need a velocity of 8 ips at the end of our 1-second maneuver, if we are to perform the desired movement of 4 inches in one second.

Note then, that it is thus an overall velocity of 4 ips which achieves your 4 inches of movement in one second, which therefore requires that we accelerate m to 8 ips at the end of our second, if the movement is to be performed at the average of the 0-to-8-ips range in velocity which will move m 4 inches in our requisite second.

So, the rate of acceleration and ultimate force we are looking for is a capacity to accelerate 1 Kg to a velocity of 8 inches per second in one second.

Thus, where a consistent uf is applied, this produces a linear rate of acceleration (as does the acceleration of gravity); and so, where the required static velocity for moving m however much we intend to move m is sv, the ultimate velocity, uv, to which we must accelerate m over that distance is 2sv, and therefore:

uv = 2sv

Thus, as 8 inches is 8/12 feet, your formula for calculating the ultimate force uf to deliver this rate of acceleration is:

uf = 1 Kg × ( ( 8 ÷ 12 ) ÷ 32 ) = 0.0208333... Kg

As we see then, the force required to execute what in fact is a relatively expedient movement of m, is far less then than m (a full Kg).

Moreover then, if we have 0.25 Kg of further resistance (Kgr), then the force required to accelerate m to an ultimate velocity of 8 ips in once second against this further resistance, must also exert the additional force of the additional resistance in order for the force we have calculated to deliver the required acceleration.

Thus our formula for the ultimate force uf to achieve the desired acceleration against the additional resistance r is:

uf = 1 Kg × ( ( 8 ÷ 12 ) ÷ 32 ) + 0.25 Kgr = 0.2708333... Kg

DETERMINING THE SUBSTANTIALLY DIFFERENT FORCES OF LIFTING m

Similarly, in the prospective act of lifting m, we observe that an upward force of m will merely sustain the position of m against the force of gravity (which is m).

Thus in regard to our formula for determining the ultimate force uf for accelerating m horizontally with no other resistance, we have the additional resistance of the mass m as acted upon by gravity to overcome, if we are to accelerate m vertically.

Thus our formula for accelerating m vertically without any other resistance becomes:

uf = m + ( m × ( v ÷ 32))

and the formula for accelerating m vertically against a further resistance fr, becomes:

uf = m + ( m × ( v ÷ 32)) + fr

Finally then, a) if m is again 1 Kg; b) our desired velocity is 8 inches per second at one second; and, c) we have an additional resistance of 0.25 Kg; then the ultimate force required to lift our object m 4 inches in one second is:

uf = 1 Kg + ( 1 Kg × ( ( 8 ÷ 12 ) ÷ 32 ) ) + 0.25 Kgfr = 1.2708333... Kg

Note that the required force to accelerate m vertically as opposed to horizontally therefore, simply increases by m.

FURTHER ACCOUNTING FOR PHASES OF ACCELERATION, CONSTANT SPEED, AND DECELERATION

Let us further consider that you may have multiple components which you need to coordinate, moving and lifting different masses to different velocities, sustained to moments at which, similarly, certain rates of deceleration may be imposed.

Here, each movement is divided into as many as 3 distinct parts. In the first, a uniform rate of acceleration may be imposed. In the second, the resultant velocity is sustained. And in the third, a necessary rate of deceleration may be applied to reduce the velocity to zero at some exact moment of time or distance.

Here then, the same formulas simply determine the force necessary to accelerate each m to the desired velocity; and that velocity is sustained until some vital moment at which the rate of deceleration must be imposed.

To determine this vital moment, we simply apply the required rate of deceleration to the resultant velocity to determine the timespan over which the rate of deceleration must be applied, to determine the duration of the final phase. This is subtracted from the moment at which we were to reduce speed to zero then, to determine exactly how long we need to sustain the ultimate velocity between the initial and conclusive phases of the movement.

Thus from a maximum velocity maxV and rate of deceleration in fps/s of rD, we may determine the seconds, s, of required deceleration to reduce the velocity to zero at the final juncture or moment of the movement:

s = maxV ÷ rD

"If" required mechanical coordinations ever impose the responsibility to make these determinations, you will be glad to be able to perform them.

HOW TO MAKE BEST USE OF YOUR ULTIMATE FORCE CALCULATIONS

So, whichever of these cases applies, determines the ultimate force uf which your leadscrew must exert.

Having determined uf, you can now use the leadscrew calculator links I pointed you toward to calculate the force required to turn your leadscrew. This determines the stepper power required to turn your leadscrew directly.

Should you want to evaluate the further prospective advantages of incorporating gear reduction, this will reduce your motor power requirements accordingly. Some further power will be required to overcome the additional resistance imposed by the type and quality of the gearbox. But overall, the mechanical advantage of the gear reduction ratio manifests in a similar reduction in required power. Again, do not forget that it is wise to multiply your minimum required power calculation substantially upward, to decide a power operating range that is sufficiently comfortable for an ideal motor.

Ultimately, the simplest calculations may require your motor to operate beyond margins of its best prospective performance. Failure to recognize this condition (by ensuring that required power is indeed delivered by such an achievable rate of RPM) can engender issues such as a theoretical impracticality or impossibility for achieving or sustaining a required rate of acceleration, most particularly in the upper reaches of potential RPM.

Say for example, your combined pitch, gearing, and potential RPMs make it impossible to sustain a required rate of acceleration in the upper ranges of a required mechanical maneuver. Obviously then (expecting that your leadscrew pitch already provides ideal mechanical advantages to your motor and gearbox), it will only be possible to deliver the intended acceleration at that upper range of RPM by sacrificing gear reduction and increasing motor power, so as brings the required RPM and power into a range which a sufficient motor can deliver.

Thus, only if your calculations, pitch, and gearing take care to keep the motor within useful ranges of its potential RPM and power band, will you best avoid suffering such prospective issues with a sufficiently overpowered motor, running well within the power and RPM ranges required by your highest rate of acceleration. The simple provision for this prospective condition therefore, is for your spreadsheet to calculate the maximum RPMs required by your highest required rate of movement. A quick glance at the power-band charts for the respective motor will then suffice to verify whether that data projects that the respective motor suffices to deliver upon your intentions. Alternately, your spreadsheet can incorporate the power provided by your motors at the maximum RPM required by your operations. Then the various motor-specific regions of your spreadsheet can either approve or disapprove each prospective motor automatically, while perhaps furthermore calculating the surplus power, or excess uf generated. The excess uf will in turn indicate how comfortable each prospective motor will be, delivering either the usual minimum power required, or the extreme power demanded by exceptional cases. Likewise, these calculations may be used to automatically select the best prospective motor and gear ratio for the job — in which ultimate automations, you may impose a wisely determined upper limit for useful excess power.

Make sure also therefore to read up on the differences between open and closed loop motors, because substantial performance advantages inhere to closed loop implementations; and the usual motor at this time, is instead of the open loop variety.

@Mike,

That is an absolutely outstanding response. I haven't asked the OP's question in a long time, but I'm grateful you took the time to set this out so clearly.

Many thanks

Much welcome, David.

This is a question that sooner or later, will confront practically every robotics, automation, or machine design initiative imaginable. But like all questions, it deserves a proper answer.

We're not politicians. We have to make things work.

I do regret that GC rules prevent uploading my spreadsheet template, because it eliminates practically all the work. Once you've waded thru any problem like this, a person should accumulate quite a toolbox. Perhaps I should make this answer a formal tutorial.