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Damn physics...

falias

Golden Member
Need help, I have to make an accurate (by 2 seconds to the minute) second, minute, or hour clock by myself using technology they had 500 years ago (pretty much no electricty, batteries, etc.), and I can't make a hour glass type of thing. Damn, after 30 minutes of googling I have no idea what to do. Anybody have any similar thing they did or have ideas?
 
You could also buy yourself some gears, hang a weight, and make a clock that would be extremely accurate as long as you kept the weight wound up.


EDIT: The pendulum would be much, much easier, however. How long does your clock need to keep time for, anyway? The more accuracy or duration you need, the longer you need your pendulum arm to be.
 
It can be anywhere from 15 seconds to an hour. I want to aim for 30 seconds (accurate by 1 second) to a minute (accurate by 2 seconds) if possible, does anyone know the formula I would use for the pendulum? How would I know how much to pull it back to release it so that each swing would be a second? Thanks for the help so far.
 
just an idea to consider, water dripper but use some mechanism to keep the water level in the source constant. dunno how accurate that would be though.

pendulum would be the best though but you have to design something to add kicks to it to replensih the enrgy losts
 
I made a pendulum that was accurate to 0.5sec after 30 minutes of running. I was a T shaped bar, and on either side of the T was two thumbtacks pointed down. At the bottom of the bar I put a very large weight. This was supported on two metal bars.
 
Originally posted by: TuxDave
I made a pendulum that was accurate to 0.5sec after 30 minutes of running. I was a T shaped bar, and on either side of the T was two thumbtacks pointed down. At the bottom of the bar I put a very large weight. This was supported on two metal bars.

Thumbtacks were definitely around 500 years ago....
 
Originally posted by: esun
T = 2*pi*sqrt(L/g)

That means the longer the length, the longer the period. Set T = 1 s, so it should be:

L = g/(4*pi^2) ~ 0.2482 meters

to original poster: that's an approximate for small angles of defelection so if you use that formula don't swing it too much
 
Yep, Pendulum is what you want. The formulas have already been posted so I won't bother with typing them in for ya.

Wait a minute, dighn's right, those are just approximations. I once had a full description of a pendulum in some lab documents, but a colleague has them, sorry.
 
what about those japanese garden things, that fill using water and empty at a set rate, very constant if you ask me, i'm sure those were around
 
Originally posted by: DrumminBoy
how about a sundial 😛

Originally posted by: Mingon
sundial 😀

Sorry, no sundial, looks like the pendulum is going to be it.

Originally posted by: dighn
Originally posted by: esun
T = 2*pi*sqrt(L/g)

That means the longer the length, the longer the period. Set T = 1 s, so it should be:

L = g/(4*pi^2) ~ 0.2482 meters

to original poster: that's an approximate for small angles of defelection so if you use that formula don't swing it too much

Do I just keep testing it to get one second for each swing, then mark that releasing point on a backboard, or is there a mathamatical way to figure that out.

Thanks for all the help guys. :beer:
 
I used a water dripper in a physics contest and nailed the times I had to measure to within .5 second on a 42 second interval. I won 🙂
 
The equation clearly shows there is no dependence on mass or deflection (although smaller deflection is better because of ignored factors). Just get a piece of string and hang it from a peg so it drops exactly 0.2482 meters (or as close as you can get to that) with a mass at the bottom (preferably something moderately heavy but small), then swing away.
 
Originally posted by: silverpig
I used a water dripper in a physics contest and nailed the times I had to measure to within .5 second on a 42 second interval. I won 🙂

Unfortunatly I can't use anything representing a hour glass with sand or water, would be so much easier.
 
Originally posted by: esun
The equation clearly shows there is no dependence on mass or deflection (although smaller deflection is better because of ignored factors). Just get a piece of string and hang it from a peg so it drops exactly 0.2482 meters (or as close as you can get to that) with a mass at the bottom (preferably something moderately heavy but small), then swing away.

So you're saying speed is not a factor, or isn't that the acceleration of gravity (g)? I would have though where you release it for it to swing would matter how fast it completes a full swing, can't just swing away.....or am I wrong?
 
Do I just keep testing it to get one second for each swing, then mark that releasing point on a backboard, or is there a mathamatical way to figure that out.

The great thing about pendulums is that the period is the same regardless of where you release it. The only thing affecting period is the length of the pendulum, so it doesn't matter where you release it.
 
Originally posted by: Venix
Do I just keep testing it to get one second for each swing, then mark that releasing point on a backboard, or is there a mathamatical way to figure that out.

The great thing about pendulums is that the period is the same regardless of where you release it. The only thing affecting period is the length of the pendulum, so it doesn't matter where you release it.

I see, so that means each swing, regardless of where it's released, would be the same lentgh of time, then wouldn't there be a standard time for one complete swing for every pendulum? How long is that?

 
I'm trying to remember the history of clocks... I'm not sure they had pendulums back then... (500 years ago)
But, I do recall that a lot of experiments were timed with water clocks...

just did a google search... pendulums started about 400 years ago.

dang it... did a search for water clocks, and the water clock site said weights and pendulums in the 14th century....
 
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