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January 28th, 2016, 02:32 PM   #11
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 Originally Posted by Country Boy If I understand this correctly, then "A" is angle, from 0 to $\displaystyle 2\pi$, giving the direction of the pointer and "B" is the angle, from 0 to $\displaystyle 2\pi$, you want to move the pointer to. So start by calculating B- A. If that is negative, replace it with $\displaystyle B- A+ 2\pi$. If it is larger than $\displaystyle 2\pi$, replace it with $\displaystyle B- A+ 2\pi$. Finally, if the final B- A, after the previous two checks, is less than $\displaystyle \pi$ rotate counter-clockwise, otherwise rotate clockwise.
Hi CB,

I have been using numbers i,e, one is more than the other, but I am interested in your method of pi and angles.

Your method looks ok, if I understand correctly, but the added complication is, that there is a DEADBAND. This means that you have to imagine the pointer being 'say' one degree each side wider, so if it is suppose to point to 90degrees, it would stop one before, depending which direction it is moving. This gives the appearance of the circle being 'say' 361 degrees around.

I hope you follow

C.

 January 29th, 2016, 12:06 AM #12 Member   Joined: Jan 2016 From: Uk Posts: 93 Thanks: 2 Hi, Here are some examples using a 0 (or 12) to 11 clock face. With a DEADBAND of '1'. So FACE (data coming in) and HAND (counting and comparing). I want the HAND to move the shortest path. 1/ HAND at 3. FACE 8. move cwise until 7 then stop. Inertia will carry the HAND farther and hopefully stop at 8. 2/HAND at 3. FACE 10. move ccwise until 11 then stop and finish at 10. 3/HAND at 0. FACE 11. don't move. 4/HAND at 0. FACE 0 don't move. 5/HAND at 11. FACE 0 don't move. The complication for me is that if FACE is 9 then move HAND till 1 before or if travelling in the other direction, 1 less. At and around 0, 1 less than 0, or 1 more than 11 (Which = 12) conflicts with 12 being 0 and starting at low again. I hope this clarifies things? I'm not a mathematician, so it completely baffles me. C.

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