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Re: Cylinder cylinder intersection: Design of a starter crank.
*To*: mathgroup at smc.vnet.net
*Subject*: [mg104847] Re: Cylinder cylinder intersection: Design of a starter crank.
*From*: dh <dh at metrohm.com>
*Date*: Thu, 12 Nov 2009 05:59:12 -0500 (EST)
*References*: <hcr7jn$8nl$1@smc.vnet.net>
Hi Matthias,
if I understand correctly, you want to calculate the intersection line
of 2 cylindrical surfaces. This can be done as follows.
Assume we have a vertical cylinder along the z axis of radius: 1. The
second cylinder is horizontal along the y axis of radius 1/2. This gives
the equations:
r1 = 1; r2 = 1/2;
eq = {
x^2 + y^2 == r1,
y^2 + z^2 == r2
};
We may solve this equations for x using e.g. Reduce:
res = Reduce[eq, {x, y, z}]
this gives:
(y == -Sqrt[1 - x^2] || y == Sqrt[1 - x^2]) &&
(z == -(Sqrt[-1 + 2*x^2]/Sqrt[2]) || z == Sqrt[-1 + 2*x^2]/Sqrt[2])
You see, there are 2 possibilities for y and two for z, giving a total
of 4 functions. We may Plot this functions using ParametricPlot. For
convenience, we change the result res from equations to rules using:
ToRules:
ParametricPlot3D[{x, y, z} /. {ToRules[res]}, {x, 0, 1},
AspectRatio -> Automatic, AxesLabel -> {"x", "y", "z"}]
The four curves join smoothly as expected.
For wall thickness you would evaluate the curves at specific points.
E.g. the maximal z value is optained from x=1 and z=Sqrt[-1 +
2*x^2]/Sqrt[2]. The min. x value from: z==0 and z=Sqrt[-1 + 2*x^2]/Sqrt[2].
Daniel
Matthias Bode wrote:
> Dear Groupmembers:
>
> I have made futile attempts at designing a "simple" starter crank to start an internal combustion engine.
>
> The starter crank has a short transversal rod that pierces the crank perpendicularly and engages in the starter ratchet mounted on the engine's cranks haft. The crank's longitudinal axis intersects the rod's longitudinal axis.
>
> The crank's lever measures 320 mm in length, the transversal rod measures 40 mm in length which is equal to ratchet's outer diameter.
>
> The diameter of the crank must not exceed 18 mm (min. 16 mm), the diameter of the transversal rod must not exceed 9 mm (min. 6 mm). - (A thinner crank disengages more easily from the ratchet when the engine starts.)
>
> I estimate the maximum force to be exerted on the crank's handle (just before the engine's crankshaft will start to rotate) to be ~ 700 N.
>
> I studied http://mathworld.wolfram.com/SteinmetzSolid.html to find out how much material of the crank would be left given the bore for the transversal rod.
>
> I got completely stuck at equation (12) - which I originally intended to solve using MATHEMATICA.
>
> Given the tensile strengths of the crank and the rod (some high tensile str ength steels) the task is to determine the diameter of the transversal rod in such a way that crank and rod will withstand about the same forces. - Obviously a very thin rod will yield to the shear strain whilst too thick a rod will leave too little material of the crank, and alter the lever ratio rod to crank to the detriment of the latter.
>
> (I already made such a starter crank using mild steel: The rod (diameter 6 mm) was shorn off AND the transversal hole in the crank (diameter 16 mm) be came more or less elliptic. - Torsion and bend of the crank itself are minimal and pose no problem.)
>
> Solutions or references to (engineering) litterature will be highly appreciated.
>
> Best regards,
>
> MATTHIAS BODE
> COCHABAMBA/BOLIVIA
> S 17.35775=B0, W 066.14577=B0
>
>
>
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