# Tag Info

20

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 ...

10

Check out my answer to a similar question at the Robotics SE. You are asking only about a motor and a wheel, and you do not give the weight of the vehicle (assuming there is one), so I cannot comment regarding vehicular motion. Basically, if you assume friction is negligible, the torque a motor outputs is used to accelerate a load. Then you have to ask, "...

9

That figure is about right for a low tensile bolt. See also this calculator and this table As a reality check if we approximate to a cross sectional area of 7 mm2 and a load of 1000 N that gives a tensile stress of 140 MPa which is below yield even for low tensile steels. In this particular context, where torque is known, the thread pitch doesn't come ...

8

There are two different aspects to start-up torque. One is associated with the motor and the other with the system which you are trying to drive. The start-up torque of a mechanical system is the minimum torque required to get it moving from a standstill. the start-up torque of a motor is the maximum torque which it is capable of producing from a ...

8

I laid everything out so you should only need to read it from top to bottom and look backwards for variables, never forward. I also tried to lay it out so hopefully you know where everything is coming from (as long as you have a basic understanding of power, torque, force, and friction...maybe even if you don't). $\mu_{roll}$ = coefficient of rolling ...

7

if you have them in line like the following picture then NO. The first and the last gear will have the same speed and the same torque (actually slightly less due to losses). if you wanted to retain a mechanical advantage you'd need 3 shafts and at least 4 gears like in the following image.

6

Detent torque is the torque while coils are not energised. Permanent magnets and the iron stator hold the motor in a given position without any power needing to be applied, although this force is usually much lower than the full torque - in my experience about 10%.

6

No. Your basic problem is that "torque is proportional to radius" is wrong. Torque is the expression of normalized rotational force independent of radius. For example if a motor can deliver 10 Nm torque, then a pulley with 1 m radius can produce 10 N of force, at 500 mm radius 20 N, at 2 m radius 5 N, etc.

6

A given mass of copper and iron can only produce so much torque : increasing torque further would require more current which means (a) the increased magnetic flux saturates the iron and (b) the I^2R losses in the copper overheat the motor. So more torque requires more iron and copper; i.e. a bigger, heavier, more expensive motor. Furthermore, because ...

6

To put it simply your loads are causing the beam/bar to bend right? So at the cut we need to consider the internal bending moment ($M$). Like you said: the professor has cut the bar to calculate the INTERNAL forces experienced by the member. In any 2D problem (say in the x-y plane) the first thing that you learn is that whenever you take a cut there's three ...

6

"Starting torque" and "locked-rotor torque" are the same thing. So if you run across the later term, you can treat it as a synonym for "starting torque." The NEMA MG-1 definition of locked-rotor torque is: "the minimum torque which [a motor] will develop at rest for all angular positions of the rotor, with rated voltage applied at rated frequency."

6

Is the torque required to make something that is not turning start turning. You can think of starting torque as an instantaneous value (the torque just before the motor starts moving). You can also thing as the torque function from that moment until the motor reaches the desired speed. The torque provided by a motor is T_{motor}(0) = T_{static friction}\\... 6 Electric motors typically only operate well at high speeds, and compared to something like an internal combustion engine, relatively low torque. Fixed gear ratios are used to cope with this inherent limitation. Additionally, some designs will use a gear train so that they can design a point of failure in their system that is not the motor. If the system, ... 6 But I'm thinking it may be better (more efficient maybe? less vibration maybe?) to run it in three-phase mode, with the load balanced among the three hot terminals. Correct. Figure 1. The load is constant through the generator cycle. Source: T. Davies - website not found. My thinking, based on my coarse understanding of generators, is that in three ... 5 Initially Assume an ideal motor with no mechanical losses and Operation in a perfect vacuum. Call the two portions rotor and stator - with all attached parts such as controller and power pack forming part of one or the other. Start at rest relative to a "fixed" frame of reference. Rotor and stator will rotate in opposite directions with the '... 5 Apart from frictional losses, which should be fairly low for a 1:4 ratio, the torque/speed ratio is inescapable ie if you increase the angular speed by a factor of 4 the torque must be reduced by a factor of 4. Power is torque x angular velocity so if this was not the case you would get more power out than you put in, which as we know is not possible. ... 4 The stator and the rotor would rotate relative to each other in the same way as if one end were attached to the Earth. If your frame of reference was attached to either the stator or the rotor, you would see the things attached to the other part rotating in the same way as if you were standing on Earth. The only difference is that, in space, you would need ... 4 First, kg-cm is not a unit of torque. On the surface of the earth, your 10 kg-cm would actually be 98 N-cm, or 0.98 Nm. If these gears are on a satellite in space, then kg-cm for torque is completely nonsensical. For sake of example, I'll assume you are relating mass to force by gravity here on the surface of the earth. Please be more careful with units ... 4 The twist introduces a sideways torque, making the belt align non-uniformly, stretch and compress non-uniformly, and rub against any stops that prevent it from falling off (or be prone to fall off in absence of such stops). This increases load and wear. If you stick to manufacturer-specified misalignment, you can fully expect the belt's life time will be as ... 4 If you want a quick and somewhat inaccurate hill-billy engineering answer, I would take a digital kitchen scale and place the PSV ontop of it, tare it, then load up a program to test the buttons (game, menu, etc). with the scale device zero'd out, slowly depress the button until it is read by the program. a full actuation can and will differ from the ... 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 The simplest way by far is to select a prospective bolt, calculate the clamping force this creates when it is torqued to rated torque, and use the bolt circle as an approximation of the moment arm. Compare this against the rated torque spec and duty type for the gearbox. Grade 5 bolts are ordinarily adequate for industrial uses. Some fussier machines or ... 4 to triple the output of a 5.5HP engine, you must either triple the displacement or triple the rotating speed. there is no way to triple the displacement by boring out the block and fitting a bigger piston and/or by installing a crank with a longer throw to increase the stroke n that engine. Now, tripling the rotating speed requires a long list of ... 4 You need to hold the shaft (in a vise with protective jaws) and turn that nut with a "C" spanner - which looks like a C with a single protruding tooth. A poor method is to use a drift or punch, but this always damages the nut - perhaps fine for a temporary repair... Edit as Fred says that is a circlip, actually in internal circlip which means that the nut ... 4 The bicycle disk as shown is inherently prone to shimmying and spiraling into an increasing flutter. I have an Ebike with a very similar disk brake and I have learned my lesson to be cautious with it. As for controlled caliper force, one could use a scissor mechanism with a hook on the bottom which could be loaded by the desired weights as per the schematic ... 3 Torque of a motor is directly proportional to current. Therefore, the maximum torque a motor can produce is limited by the maximum current it can handle. The maximum steady state current a motor can handle is limited by the heat it can safely dissipate, which is proportional to the square of the current. This means you have to be careful with that spec. ... 3 Torque ratings for bevel POM (also called acetal or Delrin) gears are at http://www.huco.com/products.asp?p=true&cat=285 I didn't find timbre boards on google, I assume you meant timber wood boards. I don't think it is feasible with wood boards screwed together to get the necessary accuracy to holding the center lines of the gears to each other for ... 3 There are various reasons why one might chose to use a geared motor with a fixed gear ratio for a robot arm over an ungeared motor: As Carlton commented on your question, it may be difficult to find an ungeared motor that meets the specific speed/torque requirements for a given application and a motor that can provide enough torque ungeared may be too large/... 3 $$\tau_{\mbox{required}} = \mbox{acceleration torque} + \mbox{friction losses} + \mbox{slope/hill climbing opposing force} \\ \tau_{\mbox{required}} = I\alpha + \mu F_{\mbox{normal}} r_{\mbox{effective}} + mg \sin{\theta} r_{\mbox{effective}} \\$$ where:\tau_{\mbox{required}}$is the required torque$I$is the load inertia$\alpha\$ is the desired peak ...

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