andyboy1970 Posted February 3, 2014 Share Posted February 3, 2014 I am looking to replace the motor in my mount (Tal 2M), the worm that drives the RA gear takes 8 minutes for one revolution ( the same as HEQ5's and 6's ) the input cog on worm has 30 teeth and my motor output turns at 0.86 revs/minute.Question is what size cog would I need on my motor output to turn the RA at the correct rate?Or would I need a combination of gears in a gearbox in-between?Cheers,Andy Link to comment Share on other sites More sharing options...
+Graham Posted February 4, 2014 Share Posted February 4, 2014 I am looking to replace the motor in my mount (Tal 2M), the worm that drives the RA gear takes 8 minutes for one revolution ( the same as HEQ5's and 6's ) the input cog on worm has 30 teeth and my motor output turns at 0.86 revs/minute. Question is what size cog would I need on my motor output to turn the RA at the correct rate? Or would I need a combination of gears in a gearbox in-between? Cheers, Andy Morning Andy Sorry but I cannot quite get my head around what you are asking. If you are replacing the motor has the motor you are removing not got a gear already on it ?? Some pics would be helpful. thanks Graham Link to comment Share on other sites More sharing options...
+Graham Posted February 4, 2014 Share Posted February 4, 2014 Morning Andy Sorry but I cannot quite get my head around what you are asking. If you are replacing the motor has the motor you are removing not got a gear already on it ?? Some pics would be helpful. thanks Graham Further to my last. If you are saying the new motor turns at 0.86 rpm then you are in trouble as that means if my sums are correct the drive gear needs to have 3.2 teeth. Link to comment Share on other sites More sharing options...
Tweedledee Posted February 4, 2014 Share Posted February 4, 2014 (edited) I have no experience with gearing calcs but after a lot of head scratching, this is how I think the problem should be solved...Taking the simple gear formula... S1 x T1 = S2 x T2Where,S1 is the motor speed = 0.86 rpmT1 is the no. of teeth on the motor cog = ?S2 is the speed required = 0.125 rpmT2 is the no. of teeth on the worm cog = 30We rearrange the formula to getS2 x T2 / S1 = T1We then get...0.125 x 30 / 0.86 = about a 1:4.3 reduction ratio or exactly 375/86You can achieve this ratio using two cogs of 375 and 86 teeth respectively.I suggest you get some confirmation of this from someone more familiar with messing with gears ratios. Edited February 4, 2014 by petersull Link to comment Share on other sites More sharing options...
+Graham Posted February 4, 2014 Share Posted February 4, 2014 I have no experience with gearing calcs but after a lot of head scratching, this is how I think the problem should be solved... Taking the simple gear formula... S1 x T1 = S2 x T2 Where, S1 is the motor speed = 0.86 rpm T1 is the no. of teeth on the motor cog = ? S2 is the speed required = 0.125 rpm T2 is the no. of teeth on the worm cog = 30 We rearrange the formula to get S2 x T2 / S1 = T1 We then get... 0.125 x 30 / 0.86 = about a 1:4.3 reduction ratio or exactly 375/86 You can achieve this ratio using two cogs of 375 and 86 teeth respectively. I suggest you get some confirmation of this from someone more familiar with messing with gears ratios. Hi Pete Just rechecked my calcs and you are correct gear needs to have 4.36. teeth. This can be achieved by changing the 30 tooth gear to a 52 and driving it with a 12 tooth gear on the motor. = 4.333333333 Which is close enough. Soon sorted with a belt drive mod as both gears are available off the shelf. Link to comment Share on other sites More sharing options...
andyboy1970 Posted February 4, 2014 Author Share Posted February 4, 2014 Cheers guys but I made an error in the original post. 48 teeth on the worm cog. So would - 0.125 x 48 / 0.86 = 6.98 be correct. Link to comment Share on other sites More sharing options...
+Graham Posted February 4, 2014 Share Posted February 4, 2014 (edited) q Edited February 4, 2014 by Graham Link to comment Share on other sites More sharing options...
+Graham Posted February 4, 2014 Share Posted February 4, 2014 (edited) try that again On the new figures. If you put a 10 tooth gear on the motor and keep the 30 tooth gear you will need to slow the motor down to 0.6 rpm -- I think Edited February 4, 2014 by Graham Link to comment Share on other sites More sharing options...
andyboy1970 Posted February 4, 2014 Author Share Posted February 4, 2014 Mmmm, scratches head! Will keep you posted. Link to comment Share on other sites More sharing options...
Big Al Posted February 4, 2014 Share Posted February 4, 2014 Andy, I've never had occasions to take a scope mount apart but I do work with motor gearbox units a lot. The principle of a worm wheel box is to give a large reduction in RPM between the motor output and the gearbox output without adding intermediate gears (reducing backlash). So the figure of 0.86 RPM seems VERY slow for a motor output?? Normally, a worm is driven direct by the motor output, so 1 rev of the worm moves the driven gear by 1 tooth, so a 30T driven gear is a 30:1 reduction. I'd guess that (because of the fractional RPM output needed) there would be more intermediate gears involved before the worm? I could be (and frequently do) talk rubbish, as the stuff I deal with is considerably bigger with 400v AC drives, though... Alan Link to comment Share on other sites More sharing options...
+Graham Posted February 4, 2014 Share Posted February 4, 2014 (edited) Andy, I've never had occasions to take a scope mount apart but I do work with motor gearbox units a lot. The principle of a worm wheel box is to give a large reduction in RPM between the motor output and the gearbox output without adding intermediate gears (reducing backlash). So the figure of 0.86 RPM seems VERY slow for a motor output?? Normally, a worm is driven direct by the motor output, so 1 rev of the worm moves the driven gear by 1 tooth, so a 30T driven gear is a 30:1 reduction. I'd guess that (because of the fractional RPM output needed) there would be more intermediate gears involved before the worm? I could be (and frequently do) talk rubbish, as the stuff I deal with is considerably bigger with 400v AC drives, though... Alan Alan What you have to remember is if you left the mount tracking at normal sidereal it would take 24 hours for the scope to do one revolution. The worm gear has a ratio of 180 : 1 if it is the same as the EQ6 Edited February 4, 2014 by Graham Link to comment Share on other sites More sharing options...
andyboy1970 Posted February 4, 2014 Author Share Posted February 4, 2014 Andy, I've never had occasions to take a scope mount apart but I do work with motor gearbox units a lot. The principle of a worm wheel box is to give a large reduction in RPM between the motor output and the gearbox output without adding intermediate gears (reducing backlash). So the figure of 0.86 RPM seems VERY slow for a motor output?? Normally, a worm is driven direct by the motor output, so 1 rev of the worm moves the driven gear by 1 tooth, so a 30T driven gear is a 30:1 reduction. I'd guess that (because of the fractional RPM output needed) there would be more intermediate gears involved before the worm? I could be (and frequently do) talk rubbish, as the stuff I deal with is considerably bigger with 400v AC drives, though... Alan Hi Alan, My figures have been revised, here goes - 1) The motor/gearbox output turns at 0.2 RPM 2) The input gear cog on the worm shaft has 48 teeth. So a cog to go between the motor output and worm input by my calculations should be 30 teeth therefore giving one worm rotation every 8 minutes. 8 minutes the motor output travels 1.6 turns (0.2 x 8 mins = 1.6) 30 teeth x 1.6 turns = 48 giving a single turn of the worm drive. Link to comment Share on other sites More sharing options...
andyboy1970 Posted February 4, 2014 Author Share Posted February 4, 2014 Now all I need is a matching 30 tooth cog to mesh with the 48 tooth cog. Found one here - http://www.bearingboys.co.uk/0_5_Mod_x_30_Tooth_Metric_Spur_Gear_In_Steel-63731-p But not sure it will match. Link to comment Share on other sites More sharing options...
andyboy1970 Posted February 4, 2014 Author Share Posted February 4, 2014 Morning Andy Sorry but I cannot quite get my head around what you are asking. If you are replacing the motor has the motor you are removing not got a gear already on it ?? Some pics would be helpful. thanks Graham Replacing the motor and electrics al-together, don't want 12vdc 50hz motor as the inverter/converters are expensive. Link to comment Share on other sites More sharing options...
+Graham Posted February 4, 2014 Share Posted February 4, 2014 Andy Lets get these figures straight. Are you now saying the motor turns at 0.2 rpm ???? Link to comment Share on other sites More sharing options...
andyboy1970 Posted February 4, 2014 Author Share Posted February 4, 2014 Andy Lets get these figures straight. Are you now saying the motor turns at 0.2 rpm ???? Yes, the motor with attached gearbox assembly, I timed it this afternoon, 1 rev = 5 minutes. I was told originally it was 1 rev every 70 seconds. So I need to make that into 1 rev every 8 minutes at the worm, this is where the 30/48 gears come in to play. Link to comment Share on other sites More sharing options...
+Graham Posted February 4, 2014 Share Posted February 4, 2014 No probs Andy You can get all the stuff you need here http://www.motionco.co.uk/timing-pulleys-timing-pulleys-aluminium-c-25_35_50.html You might as well make it belt driven whilst you are at it Link to comment Share on other sites More sharing options...
andyboy1970 Posted February 4, 2014 Author Share Posted February 4, 2014 No probs Andy You can get all the stuff you need here http://www.motionco.co.uk/timing-pulleys-timing-pulleys-aluminium-c-25_35_50.html You might as well make it belt driven whilst you are at it Cheers for the link. I assume that making it belt drive has no effect on the gear sizes? Link to comment Share on other sites More sharing options...
Big Al Posted February 4, 2014 Share Posted February 4, 2014 Yes, the motor with attached gearbox assembly, I timed it this afternoon, 1 rev = 5 minutes.I was told originally it was 1 rev every 70 seconds. So I need to make that into 1 rev every 8 minutes at the worm, this is where the 30/48 gears come in to play. Ah, so there is a primary reduction gearbox before the worm... I clearly have a lot to learn about mounts (retreats under rock with manuals). Alan Link to comment Share on other sites More sharing options...
+Graham Posted February 4, 2014 Share Posted February 4, 2014 (edited) Cheers for the link. I assume that making it belt drive has no effect on the gear sizes? No none at all still use the 30 and 48 toothed gears just get pulleys instead of gear cogs. They even have a bit on the site that you put in the centers and size of pulleys and it will work out the belt length for you. Edited February 4, 2014 by Graham Link to comment Share on other sites More sharing options...
andyboy1970 Posted February 4, 2014 Author Share Posted February 4, 2014 Saw that, very tempted, just need to measure the size of bore required. Link to comment Share on other sites More sharing options...
andyboy1970 Posted February 5, 2014 Author Share Posted February 5, 2014 Parts ordered, going to pimp my Tal2m to belt drive and it's only for visual use. One concern is the worm shaft diameter, it looks to be 16mm, wil bring a vernier home tomorrow. I have access to a workshop at work but doubt I will have a 16mm drill bit. Don't have access to a lathe though. Link to comment Share on other sites More sharing options...
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