Need help with drive train calculations

GregS

Registered User
Sep 4, 2008
505
19
18
Machesney Park, Illinois
Country
Region
I am working on determining the weight fall for my clock project. I am thinking I want this to be an wall mount clock. The pendulum (described here: The GregS Pendulum) is a seconds beater and I would like to keep the weights within the range of just below the seat board and just above the pendulum bob.

I am looking for corrections to this logic as I'm not sure of myself.

Drive gear ratio is 12 to 1 with a 1 inch dia winding drum.

The minute hand goes around 24 times per day.
24 * 8 days = 192 revs per 8 day period.

192 / 12 = 16 rotations of the drive drum over the 8 day period.

Drive drum is 1 inch dia. which is 3.1416" circumference.
16 * 3.1416 = 50" of weight drop over the 8 day period.

Is this correct? Is there a better way to do this?

Thank you!
Greg Schraiber
 

Allan Wolff

Moderator
NAWCC Member
Mar 17, 2005
761
114
43
Tulsa, OK
Country
Region
Your calculations look correct.
By adding a pulley to your weight, you will cut the drop in half, 25" of drop over 8 days.
Allan
 

tok-tokkie

Registered User
Nov 25, 2010
355
8
18
Cape Town, South Africa
Country
A seconds beating pendulum is about 1m long = 39 inches. Thus it seems the weight will have to be hung from a pulley. Remember the bottom of the weight is much below the pulley so, if it is not to move beyond the bottom of the pendulum, the fall needs to be reduced by the length of the weight & pulley. On the other hand you gain the length of the pendulum below the center of mass.
 

GregS

Registered User
Sep 4, 2008
505
19
18
Machesney Park, Illinois
Country
Region
Allen, thanks for the verification. It's odd how sometimes the simplest of things just don't look right. I had this moment when I feared that after all this work my clock would only run 5 days due to a simple math error.
 

GregS

Registered User
Sep 4, 2008
505
19
18
Machesney Park, Illinois
Country
Region
tok-tokkie, this morning I drew up a quick drawing and have decided that even with the pulley, I will not be able to keep the weight shell within the bounds noted previously. I guess that's ok as I've never seen a clock that did. By the way, your write up on your gravity escapement was a huge help to me in my design journey. I, like you, believe I have some truely novel improvements to this escapement, but keep reminding myself that it is really hard to be 'novel' after 300 years of clock design work.
 

tok-tokkie

Registered User
Nov 25, 2010
355
8
18
Cape Town, South Africa
Country
Gregg I am pleased that what I wrote of of use to you.

• My clock has 1 sec pendulum which is hung high in the case so as to reduce the overall height of the case. I also wanted to limit the fall of the drive weight so that the bottom of the weight does not go past the bottom of the pendulum. I use an endless Huygens drive.

The picture shows the drive weight on the right & the tension weight on the left close to the pendulum bob. About 1.5 days through the 8 day fall.
I use a industrial drive belt which I have perforated at 6mm pitch. The second picture, taken when the clock was at an early stage of development, shows the green belt. The central sprocket is on the hour arbor, so the clock is driven directly on the hour arbor. The two upper pulleys carry the belt across from the drive side on the right to the slack side on the left. There is a simple ratchet on one of those pulleys.

The drive sprocket has 14 teeth. The other sprockets have 18 teeth. The diameter of the drive sprocket is 25.3. That was determined by successively reducing the diameter until the pins slid nicely into the holes in the belt. In 24 hours the belt falls 14*2=28 pitches 28*6= 168mm. So in 7 days 1176 mm pass over the drive pulley. Because of the pulley on the drive weight it falls 588mm.

Most books and internet discussion about clock drive weight only considers the drive weight. The lower the drive weight the better the clock is the general opinion. This overlooks the other two factors – distance fallen and time taken.
• The weight is 1.9kg but it is counteracted by the tension weight of 200 g so the effective weight is just 1.7kg. The energy used in a week is : E=mgh
=1.7*9.81*0.588
= 9.8 Joules.
That is based on the usually accepted value of 9.81 m/s*s for gravitational attraction.
• What is really needed is the power required to drive the clock. Power is energy divided by time:
P=E/t
In this case the time is 7 days = 7*24*60*60 = 604800 seconds
So the power = 9.8/604800
= 0.000016 W
= 16 µW (microWatts)

2 Mika.JPG IMG_8356.JPG
 

Forum statistics

Threads
176,157
Messages
1,541,839
Members
53,159
Latest member
Ellaurin
Encyclopedia Pages
1,065
Total wiki contributions
3,031
Last update
-