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Weight balance formula?

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#1 wrvond

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Posted 26 October 2020 - 08:46 PM

Years ago I had a vasectomy so I can no longer multiply. Because of this I need help from a math wizard, please.  wink.gif

 

My Newt is in balance, but I want to add two or three pounds in the focuser. What formula would I use to indicate how much weight in magnets I need to add to the opposite end to keep everything in equalibrium?

 

Distance from the focuser to pivot is 22 inches. Distance from the pivot to the "magnet zone" is 16 inches.

 

Thanks!

 


Edited by wrvond, 27 October 2020 - 02:27 AM.

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#2 AstroBrett

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Posted 26 October 2020 - 08:56 PM

It is called the "lever rule. The numerical value of the weight times the distance from the pivot must be equal for both ends of the scope.

 

Feel free to send me the particulars in a message if you want me to calculate it for you.

 

Brett


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#3 Simcal

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Posted 26 October 2020 - 08:58 PM

x lbs=(3 lbs*22 in)/16 in = 4.13 lbs    (for the 3 lb scenario)


Edited by Simcal, 26 October 2020 - 09:24 PM.

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#4 wrvond

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Posted 26 October 2020 - 09:17 PM

Thanks guys, I’m good to go now!  waytogo.gif



#5 Simcal

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Posted 26 October 2020 - 09:24 PM

Nope, I did that wrong.  Fixed. foreheadslap.gif


Edited by Simcal, 26 October 2020 - 09:26 PM.

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#6 JamesMStephens

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Posted 26 October 2020 - 10:25 PM

Is this a common side effect of the vasectomy?



#7 Jon Isaacs

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Posted 26 October 2020 - 10:43 PM

Is this a common side effect of the vasectomy?

 

That's the point of a vasectomy... :)

 

Put the weight on the backside of the scope.

 

4618259-Balancing a DOB.jpg
 
Jon

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#8 Asbytec

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Posted 27 October 2020 - 05:50 AM

 

That's the point of a vasectomy... smile.gif

 

Put the weight on the backside of the scope.

 

 
 
Jon

 

waytogo.gif

 

In my experience and opinion, the problem is more than simply balance. It's about controlling the center of gravity. As the tube rotates in altitude, a displaced center of gravity relative to the horizontal altitude axis will cause the scope to move uncommanded toward the zenith or horizon, depending on where the COG is as weight is added or removed. It's best to pull the COG closer to the altitude axis and reduce the net lever and arm it presents. (The scope will not rotate on the z axis against it's own altitude bearings, so only x and y COG position need be considered). 

 

IMO, the best way to do that is to place the counter weight on the bottom of the tube opposite the focuser, heavy eyepiece, and finder(s). Just as Jon illustrates above. I try to show the problem with the COG in my illustration below (experimenting with a belly weight, ignore it). Friction (not stiction) needs to hold the scope's displaced COG. In the illustration below, the scope would only be "balanced" when the COG is directly above the altitude axis. That's precarious, we need sufficient friction to hold the scope over a range of altitude motion. Large trunnions are nice. 

 

Dob COG.jpg


Edited by Asbytec, 27 October 2020 - 06:03 AM.

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#9 JamesMStephens

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Posted 27 October 2020 - 09:27 PM

Balancing in the vertical as well as horizontal as described by Jon is good but I'll add another consideration.  For the three pound example we had 

 

x lbs=(3 lbs*22 in)/16 in = 4.13 lbs    (for the 3 lb scenario) 

 

to balance along the tube axis.  To balance on an axis perpendicular (so the scope is balanced in the vertical position) is another "lever" problem.  How far is that three pounds from the centerline of the tube?  Let's say eight inches (this assumes a 10 inch diameter tube--5 inch radius--with the center of mass of the focusing assembly three inches outside the tube).   To balance this with the 4.13 lb weight we need to satisfy this

 

4.13*X = 3*8, or X = 24/4.13 inches = 5.8 inches

 

This seems OK with our assumption of a ten inch diameter tube.  It will probably be a little out of balance but not enough to break the static friction of the bearings.  Certainly put that counterweight on the side of the tube opposite the focuser.  


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