# Tag Info

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h11 is the tolerance range. There's a standard shaft/hole chart based on dimension/tolerance. (sorry about the images, I've searched on google but cannot find everything on the same image). h11 is applied to shafts, and H11 is for holes. With your example D4h11 is Ø48mm with tolerance of -0 to -160 microns. The referenced part can be (48-0,000) = 48mm (...

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The Lagrangian for your 1D pitch system is $\mathcal{L} := T - U$, which is $$\mathcal{L} = \frac{1}{2}(mb^2 + Ma^2)\dot{\phi}^2 + \frac{1}{2}J\dot{\theta}^2 + (Ma - mb)g\sin\phi$$ The Euler-Lagrange Equations are $$\frac{d}{dt}\bigg(\frac{\partial\mathcal{L}}{\partial \dot{q}_k}\bigg) + \frac{\partial\mathcal{L}}{\partial q_k} = Q_k$$ This gives the ...

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It feels a little unintuitive, because the part is a square tube, but you can use the function Circular pattern. Select the holes on one plane as elements to repeat, select 4 repetitions and as turning axis the axis along the tube, then the holes get placed in the desired fashion.

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The energy equation can be simplified to the two equations you've referenced, given that in the first case there is negligible work, in the second case there is negligible heat transfer, and in both cases there is negligible viscous dissipation. For example, consider a heat exchanger with negligible inlet and outlet velocities and no elevation difference ...

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You can't do it with conservation of energy alone: you need another physical principle, namely conservation of momentum in the form of the Navier-Stokes momentum equation (or, if the shear stresses are negligible, the Euler momentum equation). That conservation of momentum equation will give you an expression for the surface traction density at any given ...

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uselessly general procedure: write out equations for the kinematics, then the forces, then linearize the resulting differential equation, then transform the linearized version to frequency domain, which is your transfer function. This result will vary (i.e. is parametrized by) the range of positions. A "good" mechanism would have a transfer ...

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First of all, welcome to Stack Exchange Engineering! I think what you're looking for is how the forces work through the tool, not power transmission (which is work over time; power transmission would be more like power going through a flexible Dremel attachment). But this is a great question, and recognizing that it isn't just a simple thing puts you way ...

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Many ways of transmitting power, mechanical with a steel cable with opposing winds or hydraulics using oil or compressed air come to mind. A datasheet for a product will explain the product but not always all the details of the parts used to make it.

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