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I'm a neuroscientist PhD student working on a localization problem in the brain. Rather than explain the actual details of the problem, I'll give an abstract version of it.

Imagine a clear glass marble. A small bead of red metal is embedded inside, not at the center nor at the surface. Two circles are draw around the marble at the equator and meridian. There is also a small star drawn on the marble's surface, somewhere along the meridian between the equator and upper pole.

You tilt the sphere downward 24 degrees relative to the star (the star is at the vertex of the angle formed by tilting downward). After doing so, the red bead is located at the following coordinates relative to the star, as measured by a rigid ruler: 1.5 mm away from star along a line perpendicular to the meridian; 3.5 forward from the star along a line parallel to the meridian; and then 4 mm straight down into the marble.

Knowing these coordinates, what formula would you use to determine the new coordinates of the bead, again relative to the star, if you were to tilt the marble up or down from 24 degrees? For instance, what would the coordinates be if you tilted the marble down another 21 degrees, or tilted it back up to 0 degrees?

Note that there is no azimuthal movement, only tilt, like a head nodding.

In the diagram below, the orange arrows show the measurements made outside the marble using a rigid ruler. They are perpendicular and on the plane horizontal to the ground. The 1.5 mm arrow was difficult to draw — if you had your head against the wall, the arrow would be drawn on the wall, and the 3.5 mm arrow would be perpendicular to the wall. The blue arrow points down into the marble. The view is from the side, so the meridian is the circle, and the tilt is a clockwise turn of the circle. Degrees are for the angle of the dashed lines, the upper line being parallel to the ground. The exact radius is unknown, but it is approximately 5 mm.

enter image description here

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  • $\begingroup$ Can you add a figure depicting your measurements so that we know what you're talking about ? $\endgroup$
    – Quadrics
    Commented Nov 27, 2021 at 8:55
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    $\begingroup$ What's the ball radius? $\endgroup$
    – Patricio
    Commented Nov 27, 2021 at 9:09
  • $\begingroup$ What do you mean by "forward from the star" ? $\endgroup$
    – Quadrics
    Commented Nov 27, 2021 at 10:01
  • $\begingroup$ @Potato Thanks for your questions. Hopefully I've addressed them with what I've added. $\endgroup$
    – user90664
    Commented Nov 27, 2021 at 19:42
  • $\begingroup$ @Patricio The exact radius is unknown, but it is approximately 5 mm. $\endgroup$
    – user90664
    Commented Nov 27, 2021 at 19:42

2 Answers 2

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You have the following situation

enter image description here

The $Y$'s are related to the $X's$ as follows:

$Y_1 = \dfrac{X_1}{\cos \theta} $

$Y_2 = X_2 \cos \theta - (X_1 + X_3) \sin \theta $

$Y_3 = X_3 \cos \theta + X_2 \sin \theta - \dfrac{X_1 }{\cos \theta} + X_1 \cos \theta $

This is a linear system in the three unknowns $X_1, X_2 , X_3$, while the knowns are $Y_1, Y_2, Y_3$, as well as $\theta = 24^\circ$, so it can be solved using standard techniques (Gaussian elimination, or any other method). Once $X_1 , X_2 , X_3 $ are determined the coordinates at any other angle $\theta$ can be determined from the above equations.

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  • $\begingroup$ If I made your drawing 3-dimensional and accurate to the situation I've described, the X1 line would project perpendicularly out from my computer monitor, the X2 would be parallel to the screen and to my desk, and the X3 line would point straight down at my desk. Was that your understanding when you made these equations? I ask because I don't think the X1 measurement makes a difference in determining how the coordinates of the red bead change, yet it makes a difference in your diagrams and equations. $\endgroup$
    – user90664
    Commented Dec 2, 2021 at 19:11
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    $\begingroup$ The two circles I've drawn in my solution are projections of the sphere perpendicular to the circle of the meridian (which contains the star S). Distances $X_1, X_2, X_3$ and also $Y_1, Y_2, Y_3 $ all lie in one plane, which is the plane of the meridian circle. When you tilt the sphere you are rotating this circle about its center as shown on the right image, then shifting (translating) the circle such that the rotated star has the same position as the star before tilting. This combination of rotating then translating has the same effect as rotating about the star. $\endgroup$
    – Quadrics
    Commented Dec 2, 2021 at 22:47
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Lets assign 4mm direction as positive X-axis, 3.5mm direction as positive Y-axis and 1.5mm direction (coming out of computer screen) as positive Z-axis - with the star mark considered as origin (0,0,0)

  • Because you mention that embedded red ball is not at surface of sphere, hence, we consider x,y,z coordinate system

  • Also, you intend to rotate the big clear sphere downward (clockwise from your drawings perspective - about the z-axis) - hence, the z-coordinate does not really change even after rotation

So considering only the x and y axes in 2d for simplicity, we get the diagonal distance from origin to red ball as SQ.RT. X² + Y² = SQ.RT. 4² + 3.5² = 5.315MM and the CURRENT 2d-angle of this diagonal w.r.t. x-axis = acos (4/5.315) = 41.185°.

Now rotating the sphere in xy-plane in any direction (about the z-axis) by a given angle can be easily used to calculate the new x and y coordinates in 2d - while the z-coordinate remains the same as was originally.

Also, the diagonal length too remains constant (5.315mm) as per desired rotation about z-axis - hence, calculating any one coordinate (x or y) - would yield the other too.

Hope the above is clear and helps.

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