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Jan 7, 2021 at 16:15 vote accept James Kempton
Jan 5, 2021 at 18:19 answer added Eli timeline score: 1
Jan 5, 2021 at 17:35 comment added James Kempton Thank you for your response Eli, I see how one can get the bank angle from this. However, I do not see how it is possible to move from your expression to Young's statement of $sin(\Phi)=sin(\phi)cos(\theta)$. Do you know how to do this? I also wonder if you have any comments with my own method. I'd rather like to know why it isn't working.
Jan 5, 2021 at 15:05 comment added Eli \begin{align*} \text{you can use this }\\ &\tan(\Phi)=\frac{Y_z}{\sqrt{Y_x^2+Y_y^2}}\\ &\text{where} \\ &Y=\vec{Y}_{B2} \end{align*}
Jan 5, 2021 at 14:09 history edited Qmechanic CC BY-SA 4.0
edited title; edited tags
S Jan 5, 2021 at 14:08 history suggested Nihar Karve CC BY-SA 4.0
Changed pdf to doi webpage; changed sin and cos to \sin and \cos
Jan 5, 2021 at 13:39 review Suggested edits
S Jan 5, 2021 at 14:08
Jan 5, 2021 at 13:28 history asked James Kempton CC BY-SA 4.0