teun-lll said:
7 degrees is what i measured when taking two points very close to each other and creating a line. i did this at multible points and got a steady 7 degrees, relative to the perpendicular line coming from the center to that point. So i thought it was sufficient enough to work with.
In your pdf from #1, I now understand: R is growing from 1 = 150 mm to 2 = 300 mm in one revolution = ##2\pi## radians. In short:$$
R = R_0 \Bigl (1+{\phi\over 2\pi} \Bigr )$$ with ##R_0 = 0.15 ## m. ##\quad## Correct me if I am wrong.
Simple differentiation learns that ##\displaystyle {{dR\over d\phi} = {R_0\over 2\pi}}## .
And the graph ##\ R(\phi)\ ## of course looks trivial. It gave me the idea to replace the cam by a sawtooth, not for realization but to make it easier to understand (for me, that is ...). No more problem with ##\ N > mg\ ##

!
## F = mg \tan\alpha\ ## has to drag to the right (i.e. the wedge has to be pulled to the left) over a distance ## \ R_0 /\tan\alpha \ ## to lift the load by ##\ R_0\ ## . Work done is ##\ mgR_0\ ## (keeps me happy).
As
@jrmichler indicates, associated torque at the end is ##\ 2R_0F\ ##. But average is 25% less and I figure a flywheel can smear it out for the drive motor ?
My problem/confusion is to understand where the 7 degrees comes from. I see it in Teun's picture, I do it numerically in a messy XL sheet and find 7.2, 9.1, 7.9, 7.0, 6.4 degrees at 0, ½π, π, 1½π, 2π so I'm doing something wrong (?). It's even worse when I try to do it with symbols:
Go from ##\ \phi\ ## to ##\ \phi+d\phi \ ##: tangentially ##\ R\,d\phi\ ## and radially (see above) ##\ dR = R_0/(2\pi)\, d\phi\ ## so a tangent of $${R_0\over 2\pi R} = {R_0\over 2\pi R_0(1+\phi/(2\pi)} = {1\over 2\pi+\phi }$$meaning an angle that decreases from 9 to 4.5 degrees. Would be nice for a constant torque but I don't believe it...
Can I ping
@haruspex to put me right here...?
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