New answers tagged motors
2
I think this is a variation of the gas dynamic bearing. Such bearings have a long service life due to the fact that they do not have contact between the rotating and static parts. Air passing through special slots raises the rotating part and prevents it from contacting the stationary one.
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Just as a fast estimate we assume the following:
the only friction you swing experiences is air friction and l=2m, g=10ms^2.
that $ \ \omega =\sqrt{\frac{g}{l}}=\sqrt5=2.24rad.s^-1$
we assume you set the swing to travel an angle of 2.24/2rad=1'12rad, so its frequency of 1 second. and it travels with an average speed of 2.24*2m=4.48m/s (not correct but ...
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As you note yourself, Kv = 1/Kt. 200 Kv (RPM per volt) = ~21 rad/sec per volt. Invert this to get a torque constant of 0.04775 Nm per amp. This is what you want to know. As you calculated, at peak power, the motor will draw 41.7 amps. Multiply this by the torque constant to get about 2 Nm of torque.
The issue I see is that you are assuming that all of the ...
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1.) "... if we stretch it too far, or keep it under tension for too long without firing, it would deform and then not work anymore, am I right?
In a general sense, your thoughts are fine, but let's be more specific about the different scenarios:
(1a.) - If the spring is stretched beyond its breaking strength, it breaks and would no longer work.
(1b.) - ...
2
Just to be clear, there are external forces at work here! When you are extending the spring (or whatever), the box doesn't move downwards due to the restoring force of the ground it's sitting on. If you tried to make the box move while it was floating in zero-gee, conservation of momentum would make that impossible.
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Question 1:
If we don't pull the spring hard/far enough, when we release it it wouldn't provide much force. But if we stretch it too far, or keep it under tension for too long without firing, it would deform and then not work anymore, am I right?
You are right. The effect you are describing is called creep. most materials suffer from it. However, in this ...
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Depending on how high you want it to go, you may not even need a spring, just a solenoid with sufficient travel and weight to it.
The force generated will depend on
$F=ma$
with the $m$ of the thing moving and how fast it starts and stops moving, but this is probably difficult to determine. It would be better to run some tests.
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Let's convert 7 inches to meters, 7x0.0254 = 0.1778m
You need a motor that can produce an average speed of 0.1778m/s so it must have an acceleration of 0.1778*2= 0.3556m/s^2
Now we can calculate the force your winch or motor needs. it must lift the 10kg against gravity and accelerate it
$$F = m \alpha+ mg= 10kg*0.3556m/s^2+10kg*9.8m/s^2 =10*10.155m/s^2$$
$$...
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Kv is for back EMF. It is part of what you need to model the current, in addition to the resistive and inductive model of the motor windings. The current then gives your torque via Kt. This is all to a first approximation.
Here's a paper that describes the model, in Section II.C, equations (1-3)
https://www.researchgate.net/publication/...
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Your definition of $K_t$ seems a mistake. By definition, the torque constant is simply the slope of $T (Nm)/i(amp)$ curve of a motor, and It should be noted that the parameter $K_t$ is not related to the voltage under which the motor is operated. If you use the motor at 12VDC or 24VDC this constant will remain the same. This attribute of the motor is very ...
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