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 ?
There's not any concrete mathematics that you can really apply,
Here is a much more modern high tech way to solve this problem:
**broken link removed**
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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.
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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.
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 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.
Warpspeed said:...
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.
As for "falling at an every increasing rate" I'm not sure that is right.
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.
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.The sand above the orifice supports itself as a locked matrix, so it has minimal to no "pushing downwards" effect.
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