Sand Clock

[dragonwarrior]

hi everyone,
i often wondered myself the physics behind the sand clock..
now,imagine a 1 hour sand clock or hour glass (the sand in upper part takes an hour to settle down completely).if anyone knows the physics of the sand clock,kindly let me know the following:

1) How much is the effect of gravity on the sand particles inside the clock ?
2) Normally the sand clock is kept vertical to the ground plane.If the clock is tilted to some extent making an angle to the ground,will the sand clock measure the time as usual (1 hour) or does time vary ?

 

[Sceadwian]

There's not any concrete mathematics that you can really apply, it's for the most part a chaotic system so the timing will always be approximate, it can be likened to water flow through a drain which you should be able to find plenty of calculations for.

The time will definitely vary if it's placed at an angle, when it's vertical there's higher pressure that if it's slanted so the flow will be slower, again it follows the same basic rules of a fluid system.

 

[Mr RB]

I think it will keep similar time when at an angle as to when it is vertical, as the time is determined by the egress of the sand particles from the orifice (not from any "pressure" from above or below).

So for a given orifice size the sand particles closest to the orifice will fall out at a fairly steady rate, I think reasonably independently to the amout of sand above, at least while that sand above is of enough volume to be self supporting. All guesswork of course.

 

[Warpspeed]

1) How much is the effect of gravity on the sand particles inside the clock ?
2) Normally the sand clock is kept vertical to the ground plane.If the clock is tilted to some extent making an angle to the ground,will the sand clock measure the time as usual (1 hour) or does time vary ?

Unless you are planning to leave the Earth's surface, (?) gravity is fairly constant everywhere.
But I would expect difficulties in zero G, or on Jupiter.

And I agree with Sceadwian, the rather clever conical geometry ensures it can handle fairly dramatic angles of inclination without significant change in the "head pressure" of the overlaying sand.

In interesting device. Very easy to build and initially calibrate, and very reliable if hermetically sealed to keep out moisture.

 

[duffy]

There's not any concrete mathematics that you can really apply,

I agree. Concrete would never get through an hourglass.

 

[Sceadwian]

Concrete could get through an hour glass if powder was measured through an hour glass and into a puddle of water.

Or you could grind the conrete into a powder and just run that through... =)

 

[Warpspeed]

Here is a much more modern high tech way to solve this problem:
**broken link removed**

 

[ghostman11]

Here is a much more modern high tech way to solve this problem:
**broken link removed**

the modern version is affected by tilt!!

 

[DerStrom8]

The effect of gravity will always be the same. It is the direction of the normal force that changes depending on the angle. Theoretically, when the hourglass is vertical, the only force (besides gravity) acting on the individual grains of sand is friction. when the glass is tilted, however, the normal force also acts on the particles as well as friction. This would slow them down even more, so it would take longer for the glass to run out. In order to calculate the difference in time, you would have to figure out the force of friction on each particle of sand at the neck of the hourglass, and determine out how fast it leaves the top. You would use this calculation in the other part, in which you would use it in another formula that includes the normal force on the sand. It would be extremely difficult, but you could, theoretically, figure it out using this math:

**broken link removed**

You can get the physics information on site.

Der Strom

 

[Sceadwian]

Thank you Destrom.

 

[Mr RB]

...
Theoretically, when the hourglass is vertical, the only force (besides gravity) acting on the individual grains of sand is friction. when the glass is tilted, however, the normal force also acts on the particles as well as friction. This would slow them down even more, so it would take longer for the glass to run out.
...

That sounds logical but I think there is a lot more to it than that! For a start, the orifice is much larger than a single grain, so the majority of exiting grains don't touch the sides, they only touch each other, regardless of angles (given a smallish angle). And gravity always being vertical the grains always exit and fall vertical, moving against ech other vertically, again given a smallish angle.

And even the grains that do "touch the sides" are hard grains touching a hard curved glass surface so the total friction will be very low (especially since many grains don't touch the sides). And when vertical many grains still have friction on the curved glass sides, so the anglular change may not affect total friction as much as in a simple device.

I still think the amount the hourglass slows (given a small angle) will be nowehre near the amount expected from calcuating the force and friction.

Can we use a water analogy? If you maintained the top resevoir at an exact regulated pressure, and had a curved round glass hole at the bottom, how much would the flow rate change give a smallish angular change from vertical? I don't think much at all. It would be reasonable to think that fine sand would behave at least a little like water.

So does anyone have an hourlgass?

 

[misterT]

Can we use a water analogy? If you maintained the top resevoir at an exact regulated pressure, and had a curved round glass hole at the bottom, how much would the flow rate change give a smallish angular change from vertical? I don't think much at all. It would be reasonable to think that fine sand would behave at least a little like water.

I think a water analogy is a good simplification to this problem if the clock uses very fine grained sand. The flow rate is proportional to the pressure at the orifice and the pressure is proportional to the height of the water column (measured vertically from orifice to the "water surface"). When the clock is tilted the water column becomes shorter and pressure drops.

 

[DerStrom8]

That sounds logical but I think there is a lot more to it than that! For a start, the orifice is much larger than a single grain, so the majority of exiting grains don't touch the sides, they only touch each other, regardless of angles (given a smallish angle). And gravity always being vertical the grains always exit and fall vertical, moving against ech other vertically, again given a smallish angle.

And even the grains that do "touch the sides" are hard grains touching a hard curved glass surface so the total friction will be very low (especially since many grains don't touch the sides). And when vertical many grains still have friction on the curved glass sides, so the anglular change may not affect total friction as much as in a simple device.

I still think the amount the hourglass slows (given a small angle) will be nowehre near the amount expected from calcuating the force and friction.

Can we use a water analogy? If you maintained the top resevoir at an exact regulated pressure, and had a curved round glass hole at the bottom, how much would the flow rate change give a smallish angular change from vertical? I don't think much at all. It would be reasonable to think that fine sand would behave at least a little like water.

So does anyone have an hourlgass?

I definitely agree--there is A LOT more to it than just what I mentioned, but at the very basic level, you would use the physics of a sliding material on a slope. There are many variables that come into play, but you could, theoretically, still figure it out. I'm not sure if it's possible for the human brain to take every variable into consideration, but in theory, it's possible. Besides that, you would probably have to take into account the rate of change of the speed over time. That requires calculus differential equations, at the least.

I was not saying all you would have to do is figure out the friction and figure it out from there. At least I didn't mean to. You're right. There is a LOT more to it. I am just saying that at the very heart of this problem are those physical properties.

Der Strom

 

[Warpspeed]

I am not sure that friction from "grains sliding down a glass slope" is a good way to analyze this problem.

If you watch an hourglass empty, the level obviously falls at an ever increasing rate.
The top stays pretty flat and undisturbed during emptying almost as a fluid would.
But in the very final stages and interesting thing happens. A low angle conical crater depression forms right in the middle, and you can see the grains roll down the slope towards the middle and get "sucked" downwards.

Most of the flow seems to originate from a central unsupported core of sand falling vertically through the hole. A slight initial tilt is going to have very little effect, because there will still always be sand directly vertical over the hole.

 

[DerStrom8]

The "friction" in this problem is probably not about the sand on the side of the glass. It's the friction between each individual sand particle.

Warpspeed, that "falling at an ever-increasing rate" is where the calculus comes into play. The rate of change when the glass is vertical is what needs to be determined before you can compare it to that of a tilted glass. When it's vertical, in theory, most of the force on the individual grain of sand is on its side (scraping up against the other particles). When the glass is angled, however, normal force comes into play. The frictional force on an individual particle comes from both the side and underneath. This would put more upward force on the sand, which counteracts gravity. This would slow down the flow of the sand.

Of course, another thing that should be considered is the shape of the glass.... Not sure if someone mentioned that already or not

 

[Mr RB]

I think a water analogy is a good simplification to this problem if the clock uses very fine grained sand. The flow rate is proportional to the pressure at the orifice and the pressure is proportional to the height of the water column (measured vertically from orifice to the "water surface"). When the clock is tilted the water column becomes shorter and pressure drops.

Yes and no, yes the water analogy can be a simplification, but no you can't use the height of water to determine pressure. With the sand, the height is fairly irrelevant. The sand above the orifice supports itself as a locked matrix, so it has minimal to no "pushing downwards" effect. What determines the flow is the rate that the few "free" grains in the orifice can fall.

My water analogy was to use a constant pressure, so like the self supporting sand the contrant pressure will not affect the egress of the sand/water, so basically flow is determined by how fast the bottom sand/water can fall out the orifice.

...
Most of the flow seems to originate from a central unsupported core of sand falling vertically through the hole. A slight initial tilt is going to have very little effect, because there will still always be sand directly vertical over the hole.

Bingo. The grains at the bottom centre of the orifice must fall, and will fall based on gravity (regardless of slope). Once those grains fall the grains above and on the sides of the hourglass have nothing touching them anymore (very low friction) and will fall in to replace what was lost in the orifice.

As for "falling at an every increasing rate" I'm not sure that is right. It definitely speeds up at the end when the sand becomes a lot less self supporting and the sand column is cone shaped, but in the early stages where the bulk column of sand is held cylindrically I think the rate would be fairly constant.

 

[Warpspeed]

As for "falling at an every increasing rate" I'm not sure that is right.

Actually, what I said was:

If you watch an hourglass empty, the level obviously falls at an ever increasing rate.
The top stays pretty flat and undisturbed during emptying almost as a fluid would.

I was referring to the surface level going down at an ever faster rate, simply due to the conical hour glass shape.

That seems to have been interpreted as the volume rate of sand going through the hole falling at an ever increasing rate, which is not the meaning I intended. Sorry for not wording it a bit more clearly.

 

[Sceadwian]

The sand above the orifice supports itself as a locked matrix, so it has minimal to no "pushing downwards" effect.

After thinking about it for a second you're obviously right, how many movies have you seen that show the reality of it when a tomb with a ceiling cracks and sand starts slowly pouring in even if there's 10 tons of sand over the top if it. If it were 10 tons of water and a tiny crack in the ceiling the water would come in like a laser beam pressure washer.