Moderators: gmalivuk, Prelates, Moderators General
Mighty Jalapeno wrote:Well, I killed a homeless man. We can't all be good people.
frezik wrote:The net effect is zero, but the fuel usage at any given timeslice is lower until the hourglass runs out.
Macbi wrote:frezik wrote:The net effect is zero, but the fuel usage at any given timeslice is lower until the hourglass runs out.
The center of mass has moved back in the ship, so the ship has effectively gone less distance, so the fuel has done less work, so less fuel was used. Total.
EDIT: That might have been what you were saying. I don't really get what you mean by some bits of your post.
Yeah, it's been attempted a few times I think (not by me) and it seems like the hourglass doesn't have a chance to settle down before it runs out. The problem is you need a large enough flow to actually make the effect noticible, enough sand for it to run long enough to settle and come to a steady state flow, yet light enough that it doesn't damage the scale.Meteorswarm wrote:Can you determine this with an experiment? I think that'd be the simplest way to prove it. The sand grains probably don't weigh less than .01 grams, so just get a very precise scale, like you find in a chem lab.
The force depends on the height as well as the flow rate - a larger stream of sand would produce more impact force than a thin stream.danpilon54 wrote:Well the impact and the effect of the weightless sand would not exactly cancel, as the impact force depends on the height the sand falls, while the weightless sand effect depends only on the mass of the sand and the rate of flow, so there will be some variation.
The reason I think it has an effect is because the fall height is slowly decreasing from the bottom up - consider flow out of the top glass as Qout and flow into the bottom of the glass as Qout, and that Q is the flow rate, velocity times stream area, v*A[/i]. Now you might assume that Qin is equal to Qout, but actually, since the sand pile is growing at some velocity, vpile, Qout (the sand hitting bottom) is actually greater than Qin by vpile*A.danpilon54 wrote:Im not exactly sure how the math would work though. The growing sand pile however would have a negligible effect though because the rate the pile is growing is much slower than the speed of sand falling.
uncivlengr wrote:Thanks, that's the sort of thing I was looking for, since trying to determine it intuitively is obviously not straightforward.
One thing I noticed: you defined z as the height of the top of the sand, so dz/dt would be a constant, not zero, as the pile height is decreasing at some constant rate.
The other thing is that you said "z+y = const, dy/dt = - dz/dt"
This assumes that the sand that's falling "in limbo" is negligible, but really that's one of the important aspects. The falling volume of sand is neither in 'z' or 'y' and is constantly changing, as I alluded to in my previous post.
It's isn't constant amount, though - the pile is rising, and therefore the stream of sand shortens.ThinkerEmeritus wrote:The second point is a bit more subtle and I think I may have to do some more calculating. The falling sand can be ignored if it is a constant amount, and that is what I was doing. However, it looks to me as if the difference in the speed between the movement of the tops of the two piles due to different amounts of sand in transit ought to be a small correction to the speed of the piles themselves, so that the conclusion that the force changed a bit is not threatened.
uncivlengr wrote:Since the pile is rising at some constant velocity, the velocity of the sand flowing into the bottom pile is its velocity in reference to the hourglass plus the velocity of the pile upward. Therefore dy/dt is greater than dz/dt.
So, the height between the aperture and the pile is h(t) = 0.5 g f(t).
Why neglect that effect? Unless your hourglass is growing as sand passes through it, the amount of sand in free-fall is not constant.James Scott-Brown wrote:I think that 1 and 2 balance each other exactly---at least, if the effect of the growing sand-pile in the base of the hour glass can be neglected.
Certainly, but the question is from a theoretical standpoint if it can be mathematically proven that there is a correction at all, however small.ThinkerEmeritus wrote:Therefore I think that the contribution of the movement of the falling sand is a small correction to apparent weight of the hourglass.
ThinkerEmeritus wrote:Z_{CM}(t) = Lambda z ((z/2)+H) +Lambda y(y/2)
= (1/2)Lambda z^{2} +zH + (1/2)Lambda y^{2}
and since there is a constant amount of sand, z+y=const. Then moving Lambda to the left for the moment
(d/dt) Z_{CM}(t)/Lambda = z dz/dt + H +y dy/dt
uncivlengr wrote:Why neglect that effect?James Scott-Brown wrote:I think that 1 and 2 balance each other exactly---at least, if the effect of the growing sand-pile in the base of the hour glass can be neglected.
uncivlengr wrote: Unless your hourglass is growing as sand passes through it, the amount of sand in free-fall is not constant.
James Scott-Brown wrote:I think I have found an error in Phlip's calculation:So, the height between the aperture and the pile is h(t) = 0.5 g f(t).
As s=ut+0.5at^2, I think you mean that "the height between the aperture and the pile is h(t) = 0.5 g [f(t)]^2"
phlip wrote:Were I not lazy, the next step would be to expand v and h'(t) in terms of f(t) (or in terms of h(t), which would be less fun with the square root, but probably more useful) but I am, so I won't.
Ended wrote:ThinkerEmeritus wrote:Z_{CM}(t) = Lambda z ((z/2)+H) +Lambda y(y/2)
= (1/2)Lambda z^{2} +zH + (1/2)Lambda y^{2}
and since there is a constant amount of sand, z+y=const. Then moving Lambda to the left for the moment
(d/dt) Z_{CM}(t)/Lambda = z dz/dt + H +y dy/dt
Should the RHS not be
z.dz/dt + H.dz/dt + y.dy/dt ?
Fallible wrote:1. At some stage it must accelerate downwards
2. At some stage it must reach a maximum downwards velocity
3. At some stage it must decelerate.
ThomasS wrote:If, instead of perfection, we assume that pressure caused by the sand in the top chamber affects the flow rate, then we jump from stage 1 to stage 3 when the sand starts to hit the bottom.
Wikipedia wrote:In the physical sciences, weight is a measurement of the gravitational force acting on an object.
'Wikipedia wrote:In science and technology "weight" has primarily meant a force due to gravity
Fallible wrote:ThomasS wrote:If, instead of perfection, we assume that pressure caused by the sand in the top chamber affects the flow rate, then we jump from stage 1 to stage 3 when the sand starts to hit the bottom.
Of course, the devil of this question is all in the asusmptions. However I think we can dismiss the problem of the pressure in the top chamber effecting the low rate. The reason I believe this is valid, is that hourglasses are useful precisely because the flow rate is {edit: mostly} independent of the amount of remaining sand in the upper chamber.
A A Mills, S Day and S Parkes; European Journal of Physics, 17, 97-109 (1996) wrote:The rate of flow is independent of height in the reservoir, except over the last few cm.
What we want to do is measure the 'apparent' weight of a flowing hourglass - to determine if the force between the hourglass and whatever it's sitting on changes at all. Whether or not there's any acceleration throughout the time that it's flowing is the same question, but it's more difficult to see that intuitively, because the actual container that constitutes the hourglass is not accelerating at all.Fallible wrote:All this talk about mass balances is actually irrelevant. Weight is a force, if you want to measure a force statically (as in with a mass balance), the object you are measuring cannot be accelerating!
uncivlengr wrote:The original question, since you brought it up, is if there are two hourglasses sitting on each end of a scale and one is turned over to start flowing, what will happen.
uncivlengr wrote:because the actual container that constitutes the hourglass is not accelerating at all
That won't actually affect the result. Some sand will always be accelerating and its weight will consequently be absent from the external scale. The weight of the portion of sand that reaches terminal velocity will be exerted on the scale, but its impact force will be proportionally reduced, cancelling out the effect.IHOPancake wrote:1. The sand is not in free fall. The mass of the grains of sand is small enough that air resistance should be taken into account.
In this case, we're keeping the 'container' of the hourglass at rest, but the sand isn't at rest - the scale can only measure the force applied by the hourglass system, not its mass.IHOPancake wrote:2. Weight depends only on the mass and the acceleration due to gravity. When you use a scale to measure "weight", you are actually measuring the normal force required to keep an object's acceleration at 0. [/list]
Jerry Bona wrote:The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?
IHOPancake wrote:Weight depends only on the mass and the acceleration due to gravity. When you use a scale to measure "weight", you are actually measuring the normal force required to keep an object's acceleration at 0.