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How do motorized mounts calculate sidereal time so that they can 'track'?

I've just been fiddling with Stellarium and highlighting Polaris shows no (well very little) change in the Az/Alt, but when I click futher towards the horizon the numbers start rocketing up, and on the horizon they go really fast.

I was just wondering how some mounts managed to continuously calculate this figure, based on where you are looking at the time, to keep objects in the FOV.

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I can tell you that your RA shaft needs to rotate at .000696 rpm for sidereal rate. Any motor to drive your scope would need to be geared down so that the final output will drive the scope at that rate. Gearing need not be precise, as stepper motors can be adjusted by their controllers to obtain an accurate rate. Thats if a stepper is used. Servo motors are the mode now I think.

The tech. guys will advise you more. You are contemplating wedging your Dob. aren't you.? It is not easy, and besides, you will have field rotation to contend with too. You like a dog with a rat with this, you won't let go will you. :)

RON. :)

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Alt-Az coordinates are measured from YOUR horizon and position.

The RA and DEC coordinates are measured relative to the celestial sphere.

The numbers that you see moving are Alt-Az, because the Earth is rotating, so the apparent position of the object changes too.

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But with my Alt / Az setup I dont have any use for RA and DEC stuff.

I know, I was just explaining why the numbers moved in Stellarium and the difference between the two sets of coordinates.

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I was just wondering how some mounts managed to continuously calculate this figure, based on where you are looking at the time, to keep objects in the FOV.

A motorised/computerised AltAz like NexStar 130 SLT does need to calculate the movements along the two axes. It's trigonometric calculations, if you're interested. Equatorial mounts don't have to do calculations (to follow stars) IF they are properly "polar aligned" - they can just turn at a constant rate. But they do have to adjust their rates to follow the Sun or the Moon.

I think that's correct.

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Thanks Daz :)

One more question, how much difference is there between the horizon the zenith? And how can I work that out?

Thanks Gaz for the links, im looking through them now.

trigonometric calculations

I think I'm getting a little deep here....

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One more question, how much difference is there between the horizon the zenith? And how can I work that out?

If you were on the North Pole, Polaris would be at the zenith (let's assume that Polaris is exactly on the North Celestial Pole, it's less than a degree off anyway). So, at the North Pole, things overhead would stay overhead. Things at the horizon would go nicely round, parallel to the horizon.

At the equator, Polaris would be at the horizon. Things overhead would zoom straight across the sky but things at the horizon would either do little circles (north and south) or plunge in straight lines (east and west).

Your latitude position on Earth decides what you will see. In the UK, Polaris is pretty high but not overhead. The angular distance between zenith and Polaris is (90-latitude) say about 40 degrees. I would expect the rate at the zenith to be sin(40 degrees) of the (360 degrees)/(24 hours). sin(40 degrees) is about 64%. The horizon is different because it consists of points at different angular distances from Polaris. They range from (latitude) to (180-latitude). The sin of these angles produces a factor of 77% - 100%

I think.

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If you were on the North Pole, Polaris would be at the zenith (let's assume that Polaris is exactly on the North Celestial Pole, it's less than a degree off anyway). So, at the North Pole, things overhead would stay overhead. Things at the horizon would go nicely round, parallel to the horizon.

Bet dobs are popular up there! :)

Thanks themos for the explaination, quite wordy but I sort of got the drift of it.

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There's little chance of me DIY'ing myself up a mount like the 130

Unless your able to fully understand the maths and the motions involved and

unless your proficient with electronics and accurate metal working then I would

say, with respect, that no, there's very little chance

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