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View Full Version : Wheels peripheries and planets.

John Stevenson
03-24-2005, 08:02 PM
But not in that order.

Some while ago I was involved in working out some gearing for a motorised telescope project.
I was given detals and worked a ratio out from that.

Unfortunately over the years i have lost the information.
What it was is the telescope was going to be driven by a 1 rev per minute motor anf the gearing would be worked out so that a sta or other solar object remained statinary as it was being photographed.

It was explained to me that the rotation wasn't based over 24 hours but a lesser number to take into account leap years.

Can anyone shed light on this and what would be the ratio require from a 1 rev per minute motor to keep an object stationary during the earths transit?

John S.

Rough Cutting
03-24-2005, 08:44 PM
John,
Motor turns 60 times an hour times 24 hours plus 4 turns to compensate for sidereal time. The total is 1/1444. This will make the telescope accurate to within 5 seconds a day. Hope this what you are looking for,
Gery

mklotz
03-25-2005, 11:17 AM
Hmm, I get a different answer and I'm not sure where the disagreement arises.

The sidereal day (time for one earth rotation relative to stars) is 23 hours,
56 minutes and 4.09 seconds long.

so that's:

1 rev / (23*60+56+4.09/60) min = ws (rpm)

To get this rate with a 1 rpm motor, we need a gear ratio of:

1/ws = 23*60 + 56 + 4.09/60 = 1436.06816667

which is close to but not equal to the value that Gery shows. Gery lists his
hobby as astronomy so I'm inclined to trust his value but I'd like to know what
is wrong with the above.

Regards, Marv

Home Shop Freeware - Tools for People Who Build Things
http://www.geocities.com/mklotz.geo

DennisB
03-25-2005, 12:14 PM
Some old Edmund Scientific books I have show that a 359 tooth wheel and a 4 minute worm are "exact". So, a 1436 tooth wheel and 1 minute worm should also be "exact".

This ratio gives a 23h 56m siderial day which as Marv shows isn't really "exact".

Apparently, the 4 second difference isn't significant. A long astrophoto (&gt;15 min) must usually be guided to correct for periodic error, seeing, and other anomalies.

Dennis

Paul Alciatore
03-25-2005, 01:09 PM
Marv,

Your numbers are correct. The difference is that the siderial year is exactly one revolution shorter than the sloar year due to the fact that the earth rotates about the sun once per year which adds an extra turn. The solar turns are what we call days.

So the siderial "day" is SHORTER than the solar day by the ratio of 364.25/365.25 or 0.997... When the worm wheel count of 1440 for the solar day is multiplied by that ratio you get 1436.057... Even that is not exact as the 1/4 day compensation of leap years is not exact. I'm not sure weather your number or mine is closer but both will still have an error.

Gary compensated the wrong way. His gear would work if the siderial day was longer than the solar day. But because the earth's rotation is in the same direction about it's axis and about the sun, the siderial day is shorter.

My education in physics, my personal ATM experience as well as a quick search of the ATM website's archives confirms this.

The use of a 1436:1 worm ratio is a compromise that has a sufficiently small error to allow very good viewing by eye in any telescope. However, for astrophotography, additional correction is needed to prevent smearing of the images. Traditional telescopes with "clock drives" that were used for astrophotography have always had an additional mechanism for hand correction while looking through a guide scope (astronomer's equivalent of deburring job). That is one of the reasons why you frequently see smaller scopes mounted to the side of a large one. The small scopes are also used as finder scopes as the field of view in the large scopes is so narrow that finding a particular star is difficult. And many large scopes have separate eyepieces for guideing on an off center star.

More modern scope drives frequently have stepper motors and the guideing can be accomplished via the electronics.

Paul A.

Paul Alciatore
03-25-2005, 10:04 PM
<font face="Verdana, Arial" size="2">Originally posted by Paul Alciatore:
Marv,

Your numbers are correct. The difference is that the siderial year is exactly one revolution shorter than the solar year due to the fact that the earth rotates about the sun once per year which adds an extra turn. The solar turns are what we call days.

So the siderial "day" is SHORTER than the solar day by the ratio of 364.25/365.25 or 0.997... When the worm wheel count of 1440 for the solar day is multiplied by that ratio you get 1436.057... Even that is not exact as the 1/4 day compensation of leap years is not exact. I'm not sure weather your number or mine is closer but both will still have an error.

Gary compensated the wrong way. His gear would work if the siderial day was longer than the solar day. But because the earth's rotation is in the same direction about it's axis and about the sun, the siderial day is shorter.

My education in physics, my personal ATM experience as well as a quick search of the ATM website's archives confirms this.

The use of a 1436:1 worm ratio is a compromise that has a sufficiently small error to allow very good viewing by eye in any telescope. However, for astrophotography, additional correction is needed to prevent smearing of the images. Traditional telescopes with "clock drives" that were used for astrophotography have always had an additional mechanism for hand correction while looking through a guide scope (astronomer's equivalent of deburring job). That is one of the reasons why you frequently see smaller scopes mounted to the side of a large one. The small scopes are also used as finder scopes as the field of view in the large scopes is so narrow that finding a particular star is difficult. And many large scopes have separate eyepieces for guideing on an off center star.

More modern scope drives frequently have stepper motors and the guideing can be accomplished via the electronics.

Paul A.</font>

mikem
03-27-2005, 10:39 AM
Wouldn't the apparent travel speed of the stars, planets etc. depend on where they are relative to the motion of the earth? Objects north or south would be perpendicular to the earth's travel and east and west objects appear to travel right over our heads? Thanks--Mike.