But the value shown during that process could be anywhere above or below the true weight. If you imagine just one single big grain of sand (uh, well, like a rock), when it hits the bottom of the hourglass the scale will most likely show a higher value for a short period of time.
Two more puzzles
- Thread starter Simon F
- Start date
RussSchultz said:I can't think of a way to use Archimedes principles, unless we have a twin of the chicken than we can dessicate.hupfinsgack said:Btw, anyone solved the chicken, without casualties?
edit: or if we had a block of pure "chick-onium"
Hehe.Hint: the method involves sg. also used in a CT...
24hours till I'll tell the solution, keep on riddling...
Xmas said:
I postulate a constant flow of sand!
Seriously to have 1 grain of sand you first need an opening in the upper part of the hourglass to be as small as 1 grain and secondly a flow rate equal to the rate 1 grain reaching the bottom, that's not a very practical hour glass...EDIT: And all grain must be of equal weight...
A constant flow rate of a several grains is just more practical...
Xmas said:
Constant flow ----> constant mass in the "air" -----> constant measurement mistake. However each sand grain causes displacement of the spring in the scale on hitting the bottom -----> oscillation
analogon: throwing rocks on a trampoline with the rocks remaining on the trampoline (with a constant number of rocks in the air
)Since you're from Germany it'd be a lot easier for us to talk in German via pms... However, if you want the discussion going on here in English, it's fine with me as well...
Constant flow -> constant displacement, no oscillation. It will simply find a balance. Analogy: put your scale under a water tap (water molecules = grains of sand). If you keep the water flow constant, the measured "weight" will be constant, but > 0.hupfinsgack said:Constant flow ----> constant mass in the "air" -----> constant measurement mistake. However each sand grain causes displacement of the spring in the scale on hitting the bottom -----> oscillation
analogon: throwing rocks on a trampoline with the rocks remaining on the trampoline (with a constant number of rocks in the air
Nah, got to practise my language skills
Xmas said:Constant flow -> constant displacement, no oscillation. It will simply find a balance. Analogy: put your scale under a water tap (water molecules = grains of sand). If you keep the water flow constant, the measured "weight" will be constant, but > 0.hupfinsgack said:Constant flow ----> constant mass in the "air" -----> constant measurement mistake. However each sand grain causes displacement of the spring in the scale on hitting the bottom -----> oscillation
analogon: throwing rocks on a trampoline with the rocks remaining on the trampoline (with a constant number of rocks in the air
No... sand grains are particles; constant flow refers to kg/(m²s) ----> same mass in the air; however as all grains have different size and weight, microscopically no equilibrium can be reached since the process is statistically random. It just depends on how accurate your scale is...
The answers offered would do fine.AAlcHemY said:
Another answer would be take (assuming n bags) n-1 of the bags and weigh them. It will either weigh(n-1)*(1bag of toffees) or (n-2)*(1bag of toffees)+ 1*(1bag of peanuts).
If it weighs the first, eat from those bags, if it weighs the second, eat from the bag you didn't weigh.
Though personally, if you're deathly allergic to peanuts, I wouldn't eat any of them.
With one grain, you're right Xmas.
First, think impulse: m*v = I F*dt // Read that as integral of F wrt t
If v is the same at start and end (=0), then F must be higher some time to compensate for the time it was lower.
Before the sand starts to flow, the scales show the weight of the hourglass and every grain of sand. As the sand starts to flow, the measured weight will decrease linearily until the first grain hits the bottom. Just before the grain hits the bottom the measured weight will be original weight minus the sand in the air.
The measured weight after the sand has started to hit the bottom depends on what model you use for that collision.
If the grain is modelled to stop instantly, the scales will keep showing the lower weight except for some spikes when the individual grains hit the bottom. (The average would however be full weight, including the weight of the falling sand.)
If it (more naturally) is assumed that that time it takes for a grain to "settle" after the collision is large compared to the time between two grains, the above measurent would be "low pass filtered". So you'd get the full weight (again, including the weight of the falling sand), but with a noise on top (I woudn't call it oscillation due to it's random nature).
When the last grain has started its decent, the scales will show an lineaily increasing weight. Increasing at the same rate as it was decreasing in the beginning. And when the last grain stops, the weight will jump back down to "full weght".
It's left as an exercise for the student to extend this to include the fact that as more sand gets to the bottom part, there will be a shorter way for he sand to fall, and thus less weight "in flight".
About the chicken: That's related to another thread here. You simply microwave it together with a reference glass of water. Then measure how much it heats up.
(I've already started filing all the monkey adoption papers.)
First, think impulse: m*v = I F*dt // Read that as integral of F wrt t
If v is the same at start and end (=0), then F must be higher some time to compensate for the time it was lower.
Before the sand starts to flow, the scales show the weight of the hourglass and every grain of sand. As the sand starts to flow, the measured weight will decrease linearily until the first grain hits the bottom. Just before the grain hits the bottom the measured weight will be original weight minus the sand in the air.
The measured weight after the sand has started to hit the bottom depends on what model you use for that collision.
If the grain is modelled to stop instantly, the scales will keep showing the lower weight except for some spikes when the individual grains hit the bottom. (The average would however be full weight, including the weight of the falling sand.)
If it (more naturally) is assumed that that time it takes for a grain to "settle" after the collision is large compared to the time between two grains, the above measurent would be "low pass filtered". So you'd get the full weight (again, including the weight of the falling sand), but with a noise on top (I woudn't call it oscillation due to it's random nature).
When the last grain has started its decent, the scales will show an lineaily increasing weight. Increasing at the same rate as it was decreasing in the beginning. And when the last grain stops, the weight will jump back down to "full weght".
It's left as an exercise for the student to extend this to include the fact that as more sand gets to the bottom part, there will be a shorter way for he sand to fall, and thus less weight "in flight".
About the chicken: That's related to another thread here. You simply microwave it together with a reference glass of water. Then measure how much it heats up.
(I've already started filing all the monkey adoption papers.)
Basic said:
This is second time today sb from Sweden clarifies my words
I guess I have to improve my English... About the chicken:
Congratulation! That's a new method! But the chicken is still hurting! Keep on going! Maybe we can get as many solutions as Niels Bohr had for measuring the height of a building with a barometer
As for the chicken: you need a big centrifuge.
That's why I said "related".hupfinsgack said:
I think it's more likely to look like this...The sand however falls; the amount of the falling sand is therefore not measured by the scale; however at the end when the last bit of sand the weight goes back to normal. A simple sketch would look like this:
Sketch with typo
Please excuse the typo, was done in 10 seconds.
A mor complicated model would include a function of how the sand starts to roll / fall...
If you want to go microscopic: each sand particle is going to cause an oscillation of the spring in the scale...
Code:
-(b)---
(a)------ -------
\Not without casualties - I thought maybe there might be a way based on first of measuring weight and volume (by displacement (of air not water because you can't weigh a wet chook)) getting it to drink and then repeating the process... but that doesn't seem to lead anywhere, or at least I couldn't see that you could separate matter that on average had the same density as water.Btw, anyone solved the chicken, without casualties?
I suppose you could feed it heavy water and measure the relative increase in radioactivity.
My preferred option was to connect the chook up to a fire hose and turn the tap. The % of non water matter would thus approach zero and so we can ignore it
Or, then again, is it a rubber chicken?
A new puzzle: This is one from one of my colleagues....
A large smooth circular metal ring is standing upright on the ground. Two very heavy weights, which are designed to slide around the ring, are positioned at the top of the ring. I'll try to draw a diagram:
The weights are initially held at the top and then released. What happens?
A large smooth circular metal ring is standing upright on the ground. Two very heavy weights, which are designed to slide around the ring, are positioned at the top of the ring. I'll try to draw a diagram:
Code:
-WW-
/ \
/ \
| |
\ /
\ /
----
----------Ground---I think it's more likely to look like this...The sand however falls; the amount of the falling sand is therefore not measured by the scale; however at the end when the last bit of sand the weight goes back to normal. A simple sketch would look like this:
Sketch with typo
Please excuse the typo, was done in 10 seconds.
A mor complicated model would include a function of how the sand starts to roll / fall...
If you want to go microscopic: each sand particle is going to cause an oscillation of the spring in the scale...
(A) is the steady start/end state - there'd be an initial drop and then it would measure higher at (b).Code:-(b)--- (a)------ ------- \
Sorry, but that's wrong. At the moment I am very busy... just read the post from Basic, which is correct.
Not without casualties - I thought maybe there might be a way based on first of measuring weight and volume (by displacement (of air not water because you can't weigh a wet chook)) getting it to drink and then repeating the process... but that doesn't seem to lead anywhere, or at least I couldn't see that you could separate matter that on average had the same density as water.Btw, anyone solved the chicken, without casualties?
I suppose you could feed it heavy water and measure the relative increase in radioactivity.
My preferred option was to connect the chook up to a fire hose and turn the tap. The % of non water matter would thus approach zero and so we can ignore it
That's nearly there... Take heavy water, feed it to the chicken. Wait till the chicken takes a piss. Measure the amount of heavy water (tritiumO2) and the amount of piss.
inital amount of heavy water /Water in chicken=amount of (heavy water + amount of its byproducts)/ amount of water in piss
Simon F is the winner...
Not to burst your bubble, but chickens don't piss.hupfinsgack said:
