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| - | ==Mathematical Modeling==
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| - | ===Alpha===
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| - | ====Problem====
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| - | Given the following construction and point ''p'', or (''x'',''y'',''z'') find the angles ''θ<sub>1</sub>'', ''θ<sub>2</sub>'', and ''θ<sub>3</sub>''.
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| - | Note that positive ''z'' is the down direction.
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| - | ====Constants====
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| - | *''TR''
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| - | **Top radius
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| - | *''BR''
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| - | **Bottom radius
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| - | *''L''
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| - | **Linkage
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| - | *''CA''
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| - | **Control Arm
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| - | *''φ<sub>1</sub>'' and ''φ<sub>2</sub>''
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| - | **Two angles
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| - | ====Construction====
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| - | From the top to the bottom:
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| - | *A circle centered at the origin with radius ''TR'', named ''O''
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| - | *Make three lines ''A<sub>1</sub>'', ''A<sub>2</sub>'', and ''A<sub>3</sub>'' such that:
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| - | **The lines ''A<sub>x</sub>'' are perpendicular to a tangent of ''O'' and a radius of ''O'',
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| - | **The angle between the radii of ''A<sub>1</sub>'' and ''A<sub>2</sub>'' is ''φ<sub>1</sub>'', and
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| - | **The angle between the radii of ''A<sub>2</sub>'' and ''A<sub>3</sub>'' is ''φ<sub>2</sub>''.
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| - | *Make a circle centered at point ''p'' with radius ''BR'', named ''P''
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| - | *Find three points ''P<sub>1</sub>'', ''P<sub>2</sub>'', and ''P<sub>3</sub>'' such that:
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| - | **They are on the circumference of ''P'',
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| - | **The angle between the radius which touches ''P<sub>1</sub>'' and the radius that touches ''P<sub>2</sub>'' is ''φ<sub>1</sub>'',
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| - | **The angle between the radius which touches ''P<sub>2</sub>'' and the radius that touches ''P<sub>3</sub>'' is ''φ<sub>2</sub>'',
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| - | **The ray from the center of ''P'' to ''P<sub>x</sub>'' is parallel to the ray from the center of ''O'' to the point which is on ''A<sub>x</sub>'' and ''O'' 's circumference, for all ''x''.
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| - | *Construct three line segments ''CA<sub>1</sub>'', ''CA<sub>2</sub>'', and ''CA<sub>3</sub>'' such that:
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| - | **''CA<sub>x</sub>'' is in the same plane as ''A<sub>x</sub>'', for all ''x'', and
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| - | **The angle between ''CA<sub>x</sub>'' and ''A<sub>x</sub>'' is ''θ<sub>x</sub>'', for all ''x''.
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| - | *The distance between ''P<sub>x</sub>'' and the end point of ''CA<sub>x</sub>'' that is not on ''A<sub>x</sub>'' is ''L'', for all ''x''.
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| - | ====Solution====
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| - | #Note that the control arms can only move in a circle, while linkage can move in a sphere.
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| - | #We will calculate ''θ<sub>1</sub>'' first, which only involves the points, circles and lines ''p'', ''CA<sub>1</sub>'', ''A<sub>1</sub>'', and ''P<sub>1</sub>''.
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| - | #Find the plane where ''CA<sub>1</sub>'' 's circle resides in. Use it to cut the sphere around ''P<sub>1</sub>''. For future reference, call ''CA<sub>1</sub>'' 's circle ''C<sub>1</sub>'' and the circle resulting from the cut ''C<sub>2</sub>''
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| - | #Define ''x<sub>o<sub>1</sub></sub>'', for x offset, for the difference in the ''x'' coordinates of the center of ''C<sub>1</sub>'' and the point ''P<sub>1</sub>''.
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| - | #*''x<sub>o<sub>1</sub></sub> = TR - (BR + x)''
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| - | #*''x<sub>o<sub>1</sub></sub> = TR - BR - x''
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| - | #Define ''D<sub>1</sub>'' to be the distance between the center of ''C<sub>1</sub>'' and ''C<sub>2</sub>''. It will also be the line connecting the centers.
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| - | #*''D<sub>1</sub> = sqrt(x<sub>1</sub><sup>2</sup> + z<sup>2</sup>)''
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| - | #Obviously, ''D<sub>1</sub>'', ''CA<sub>1</sub>'', and linkage form a triangle. The angle between ''CA<sub>1</sub>'' and ''D<sub>1</sub>'' is a close approximation to ''θ<sub>1</sub>'', but it is not exact. We will call this angle ''θ<sub>1<sub>1</sub></sub>''. We use the law of cosines to calculate ''θ<sub>1<sub>1</sub></sub>''.
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| - | #*''L<sup>2</sup> = CA<sup>2</sup> + D<sup>2</sup> - CA * D * cos(θ<sub>1<sub>1</sub></sub>)''<br><br>
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| - | #*''θ<sub>1<sub>1</sub></sub> = cos<sup>-1</sup>((D<sup>2</sup> + CA<sup>2</sup> - L<sup>2</sup>)/(2*D*CA))''<br><br>
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| - | #One endpoint of ''D<sub>1</sub>'' is on ''A<sub>1</sub>''. Thus, we can make another triangle, and the angle between ''D<sub>1</sub>'' and ''A<sub>1</sub>'' is the difference between ''θ<sub>1</sub>'' and ''θ<sub>1<sub>1</sub></sub>''. We will call this angle ''θ<sub>1<sub>2</sub></sub>''. If we have the length of the side on ''A<sub>1</sub>'' be ''z'', then the last side will be ''x<sub>o</sub>'' and the triangle will be a right angle triangle. Since only ''x<sub>o<sub>1</sub></sub>'' changes sign in the good interval, we should use a trigonometric function that involves ''x<sub>o</sub>'' in the numerator. Thus,
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| - | #*''θ<sub>1<sub>2</sub></sub> = sin<sup>-1</sup>(x<sub>o<sub>1</sub></sub>/D<sub>1</sub>)''<br><br>
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| - | #*''θ<sub>1</sub> = cos<sup>-1</sup>((D<sup>2</sup> + CA<sup>2</sup> - L<sup>2</sup>)/(2*D*CA)) - sin<sup>-1</sup>(x<sub>o<sub>1</sub></sub>/D<sub>1</sub>)''
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| - | #You just have to rotate the model ''φ<sub>1</sub>'' degrees counterclockwise for ''θ<sub>2</sub>'', and another ''φ<sub>2</sub>'' degrees for ''θ<sub>3</sub>'' using the two dimensional rotational matrixes
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| - | {''cos(φ<sub>2</sub>),-sin(φ<sub>2</sub>)''}
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| - | ''R<sub>1</sub>''={ }
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| - | {''sin(φ<sub>2</sub>) cos(φ<sub>2</sub>)''}
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| - | {''cos(φ<sub>3</sub>) -sin(φ<sub>3</sub>)''}
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| - | ''R<sub>2</sub>''={ }
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| - | {''sin(φ<sub>3</sub>) cos(φ<sub>3</sub>)''}
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