# Tag Info

26

You have the right concept, but slipped a decimal point. 5 cm = 0.05 m. The gravitational force on your 450 g mass is 4.4 N as you say, so the torque just to keep up with gravity is (4.4 N)(0.05 m) = 0.22 Nm. However, that is the absolute minimum torque just to keep the system in steady state. It leaves nothing for actually accellerating the mass and for ...

11

This would definitely fit under any of the following names: XY Table 2D Positioner XY Stage XY Positioner

8

Well that part's manufacturer calls it a "Precision XY Motorized Linear Stage". The terminology is not hard and fast, though. I've also heard those called "XY tables". I personally recommend that we only call them "tables" if they either: Have a relatively large platform, EG: Or the piggyback actuator is adequately supported. EG:

4

In the text you quoted, I was talking about mechanical power out of the motor. This is what the power rating of a motor, given no other qualifications, means. A "15 W motor" therefore can put out 15 W of mechanical power under the right conditions. No motor is 100% efficient, so more electrical power than that will need to be supplied to the motor. For ...

4

Yes, the holding torque goes up proportionally with the gear ratio. Plus the friction of the gear train, and any other seals, bearings along the gear train that add friction.

4

Although the advertised advantage of a stepper is that it can be driven "open loop" (no position or velocity feedback required for accuracy, instead you just keep count of the steps) this is only true when the speed with which the step signals are being fed to it are below a certain rate threshold. Beyond that threshold, the stepper armature begins to lag ...

4

Screw: M6 M6 coarse has a 1 mm pitch. M6 fine has a 0.75 mm pitch. Unless otherwise stated assume coarse, so 1 mm pitch. Your motor is 1.8° per step. This is 360/1.8 = 200 steps/rev. The screw will advance 1 mm every 200 steps. As usual with Amazon, there are no datasheets and poor specifications in the ad which is what you linked to. There's a question in ...

3

There is nothing inherently wrong with running a motor in stall. The difficulty is that the power dissipation may be higher than the motor can handle long term. Every motor can dissipate some amount of power safely indefinitely. In stall, the power being dissipated as heat is the full voltage times current. As long as that doesn't exceed what the motor ...

3

Operating principle is different. In essence if you buy just a motor it will be just that. You can now add gearing that will reduce speed and increase torque. Since torque is directly proportional to gear ratio and mechanical advantage; $$MA = \frac{v_{input}}{v_{output}}.$$ You get a hundred times more torque if you gear 3000 rpm down to 30 rpm. So now ...

3

Direct answers to question: Whether a motor could overrun depends on motor power, momentum of the load, and resistance of the system. This could be a problem even with switches, but can be overcome with careful design. Steppers are overkill if you need to move between two points without stopping between them. See below. What you need is a linear actuation ...

3

The limit on the stepper motor is the torque it can produce. This has nothing to do with the weight it is rotating, except that the weight may cause more friction, which the motor has to overcome. The mass does matter somewhat, but only in a rotational sense, and only to determine how fast the motor's torque will speed up the rotation. The proper term for ...

3

RPM of a stepper motor Formula for calculating stepping motor speed. $$RPM = \dfrac{a}{360} \cdot f_z \cdot 60$$ $RPM$ = Revolutions per minute. $a$ = step angle $f_z$ = pulse frequency in hertz

2

Without more details about your design, the dynamics of your mechanism, or your calculations, its virtually impossible to address your question with any certainty. However, i offer some advice that may prove useful to you: If affordable, I would use double the minimum holding torque needed for the application because i usually find that my calculations ...

2

You provided a lot of redundant information which is moot to the answer. Whether the pulleys drive the ring through contact, the belt, how the belt goes etc is all moot to the answer. The only thing that matters is the gear ratio between a pulley and the ring (since the motor drives the pulley directly, the gear ratio between the motor and the pulley is 1.) ...

2

Assuming a perfect transmission, the amount the pulley needs to turn for a full revolution of the large ring is governed by the "gear ratio": $\theta_{ring} = 2 \pi * 1074 / 100 = 67.48$ rad or 10.74 rev. Assuming you are full stepping, this is 2,148 steps. If you are half-stepping, it's 4,296 half-steps or if quarter-stepping, 8,592 quarter-steps. In ...

2

"Design of experiments" could help you achieve the minimum number of tests to come up with a mathematical model (transfer functions) of your setup. You can try two methods: First is the BLACK BOX approach (you have no clue what the camera is doing). You could divide the whole movable region into sections, elevationwise and azimuthwise. By doing different ...

2

One way might be to try to model and understand the physical problem. You are however clear in your question that you would prefer an empirical approach where constants are tuned to experience data. I would suggest that the easiest and most general way is to use a piece-wise approximation of the velocity curve. Generically this may be made by using basis ...

2

Depends on how much weight and the motor. Some stepper motors have detent torque. This is the torque they will hold without consuming any power. They also have a holding torque for which you need to supply power. The most fool-proof solution is to use a external brake.

2

You want a transmission with a ratio R such that $R \cdot 200 / 360$ gives an integer number of steps per degree. Then in controlling software you can program it to take e.g. 5 steps to move one degree. As Brian Drummond mentioned in a comment, 36:20 is one possible ratio, giving 1 step per degree. Some other options: 9:5, equal to 36:20 except smaller ...

2

Constrain the driven shaft and use a coupling to the motor.

2

You want something called a gearmotor, which is a modestly-sized DC motor that can run in either direction at a variety of speeds, depending on the voltage you feed it. You select the appropriate gear ratio to yield the RPM you want. See the Marlin P. Jones & Associates on-line catalog for a good selection of inexpensive gearmotors. Also note that it is ...

2

Actually if you want to keep accuracy and increase the torque you can use a gearbox that reduces the input Rpm. So you will be Sacrificing the speed for the increased torque and precision. Its also linear so if you need to go from 200gr to a kg you need a great ratio of. 5. If you had to go higher than 10 then the forces would be significant and you'd need ...

1

It is the same as everywhere else. You do not need to calculate torque based on the design of the reducer/motor. you need to figure out the ratio that specific gear vs motor will spin at. Once you do that, calculate the power of the pull, are you pulling with 5lbs or 10lbs of force? once you figure out the pull/push factors, you can translate these values ...

1

There are some "exotic" thread sizes used for small instrument work. Note that for a specific diameter there can be more than one pitch. It is also possible that the thread is not metric. Metric fine pitch threads

1

An easy solution is to use a Motorized Linear Slide Pot - they're not expensive and they have a motor and a pot - you still need to build the feedback loop to control the position and use a DAC to provide a voltage for the setpoint. A mechanical linkage can reduce the range of motion to what you require.

1

Couple the stepper motor to the high speed spindle motor only indirectly by means of magnets rotating around the spindle motor shaft which, at low speeds, will align themselves with specially designed ferrous spots fixed to the high speed spindle shaft and sync with it. The magnets should create a detent force strong enough to move the spindle at slow speeds,...

1

A one way clutch bearing on the stepper motor might give you the clutching action you want without much added mechanical complexity (note that this would require some experimentation). The high rpm motor would be attached to the tool, and would spin in the free wheeling direction. The stepper motor would be attached to the one way clutch bearing and index in ...

1

The value Detent Torque or Cogging Torque is what you should look for in the specification sheet. The detent torque is related to stepper motor power loss. The following is a reference from Minebea' web site. Detent Torque: amount of torque that the motor produces when it is not energized. No current is flowing through the windings. Here is an excerpt ...

1

cartesian 2 axis robot 2 axis robot cartesian coordinate robot xz axis robot xz gantry For your purposes, compare the required precision, the load weight, and the control board and display, pc link, if it uses GCode or some other control. Note that this kind of cartesian robot system has been popularized by 3d printers and there are masses of fan info'...

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