 Filling a water boiler at the drinking fountain sonhouse Posers and Puzzles 02 Mar '07 16:29
1. 02 Mar '07 16:292 edits
You have one of those plug in water pots, you try to fill it at the water fountain, which only shoots a stream of water up 10 Cm high and 5 Cm over. The pot is a cylinder 15 Cm high and 8 Cm diameter. How many liters max can you fill the pot just using that spout of water?
I should add the bottom of the pot is on the same level as the bottom of the water spout
BTW, this is a visualization of a real world situation where I was filling my water boiler on afforementioned water spout. I see math everywhere!
2. 02 Mar '07 17:06
Originally posted by sonhouse
You have one of those plug in water pots, you try to fill it at the water fountain, which only shoots a stream of water up 10 Cm high and 5 Cm over. The pot is a cylinder 15 Cm high and 8 Cm diameter. How many liters max can you fill the pot just using that spout of water?
I assume the cylinder must be tilted in order to catch the water, otherwise this would be fairly trivial. So, I will assume that one point of the cylinder will touch the ground at the same elevation as the origin of the water spout.

The best configuration to catch the water will be when the bottom lip of the pot just intersects with the top of the parabola formed by the flowing water. If you move the tilted pot any closer, the water will just hit the side of the pot; if you move it any further away, you'll have to tilt the pot more which will make some of the water run out.

Now, I think I'll give an overview of my solution first as I'm trying to do the math on the fly and it's slightly more complicated than I thought (but I will post my calculations too):

1. Determine the angle the pot makes with the ground, and subsequently the angle the water makes with the side of the pot on the interior (use trigonometry)

2. Tilt the pot upright, centre it at the origin on the xy-plane, and see where the plane determined by the angle of the water with the pot intersects the xy-plane (use trigonometry)

3. After determining the equation of the plane z = f(x,y), and using the fact that the equation of the cylinder in x^2 + y^2 = 16, perform a double integral with the following form:

V = int((f(x,y)))dydx, with limits of integration y = -SQRT(16-x^2)...+SQRT(16-x^2), x = -4...+4

(use calculus)

This will give you the maximum volume of water captured in the pot. Now off to calculate!
3. 03 Mar '07 00:451 edit
Do you really need to integrate? I'm thinking that the volume of a cylinder, filled with a liquid and angled such that the liquid is just touching the bottom of the cylinder on one end and just about to pour out of the cylinder on the other end, will be exactly half of the volume of the cylinder if it's standing upright. Thus, you could simply solve geometrically.

The triangle formed by the height of 10 cm and hyp. of 15 cm has bottom edge length ~11.18034 cm. The triangle formed by the "top of the cylinder" edge length of 8 cm, hyp. is top of liquid, is a similar triangle and the length of the "half-empty" cylinder is thus ~8.944272 cm. From there compute volume = pi*(0.5*8.944272+6.055728)*r^2.

Edit: ~ 529.2 cm^3 or a little over a half liter.
4. 03 Mar '07 05:12
Originally posted by leisurelysloth
Do you really need to integrate? I'm thinking that the volume of a cylinder, filled with a liquid and angled such that the liquid is just touching the bottom of the cylinder on one end and just about to pour out of the cylinder on the other end, will be exactly half of the volume of the cylinder if it's standing upright. Thus, you could simply solve ...[text shortened]... me = pi*(0.5*8.944272+6.055728)*r^2.

Edit: ~ 529.2 cm^3 or a little over a half liter.
Brilliant! And much easier. 🙂
5. 04 Mar '07 08:08
Originally posted by leisurelysloth
Do you really need to integrate? I'm thinking that the volume of a cylinder, filled with a liquid and angled such that the liquid is just touching the bottom of the cylinder on one end and just about to pour out of the cylinder on the other end, will be exactly half of the volume of the cylinder if it's standing upright. Thus, you could simply solve ...[text shortened]... me = pi*(0.5*8.944272+6.055728)*r^2.

Edit: ~ 529.2 cm^3 or a little over a half liter.
How do you know the pot is at that angle?
6. 06 Mar '07 16:16
Originally posted by AThousandYoung
How do you know the pot is at that angle?
The question referred to the max it could hold so knowing the best angle you put it there.
7. 07 Mar '07 07:231 edit
Originally posted by sonhouse
The question referred to the max it could hold so knowing the best angle you put it there.
That angle is not the best angle. The best angle is zero degrees. This would give twice the volume that the assumed angle did.
8. 07 Mar '07 07:3311 edits
You're going to need to tilt the cylinder until there are only 10 cm from the surface to the lip so the water can get in. This will give a right triangle with a 15 cm hypotenuse and a 10 cm side.

Let @ be the angle of the can to the surface.

sin @ = 10/15

@ = 41.8 degrees.

So we need to find how much water a cylinder tilted at 41.8 degrees from the horizontal, or 48.2 degrees from the normal, can hold.

While the cylinder is tilted and filled, the surface of the water will make a 41.8 degree angle with the side of the cylinder.

The surface of the water will on one side touch the lip (while the cylinder is tilted) and on the other side will be 8.9 cm down the side - NOT all the way down as was assumed!

The hard part will be to find the volume of the region left empty. This region will be triangular from the side of the cylinder, with the triangle having sides 8 cm, 8.9 cm and 12 cm, and circular from the top with a diameter of 8 cm. Not a trivial shape to find the volume of.

Edit - Ah! However if you let the can stand up again, the water will settle to a depth equal to the depth it had in the center while the cylinder was tilted...which is 15 cm minus half of 8.9 cm. So, it will be filled 10.5 cm and the volume of water will be

10.5 cm x pi x (4 cm)^2 = 527 mL

The whole can has a volume of 15 cm x 16 cm^2 x pi = 753.6 mL

which is not twice what the other guy got. However my answer and his are awfully close; just about the same considering I was rounding numbers off during the calculation.

Ah, I see...he confused me by referring to the "half empty" cylinder, and the cylinder he referred to is not the cylinder being filled but only the part that contains the surface...yeah he's right.
9. 15 Mar '07 22:38
Originally posted by sonhouse
You have one of those plug in water pots, you try to fill it at the water fountain, which only shoots a stream of water up 10 Cm high and 5 Cm over. The pot is a cylinder 15 Cm high and 8 Cm diameter. How many liters max can you fill the pot just using that spout of water?
I should add the bottom of the pot is on the same level as the bottom of the water s ...[text shortened]... ation where I was filling my water boiler on afforementioned water spout. I see math everywhere!
how high is the water spout? that would determine wether you needed to tilt the cylinder
10. 16 Mar '07 00:00
Originally posted by joe shmo
how high is the water spout? that would determine wether you needed to tilt the cylinder
11. 16 Mar '07 00:32
Originally posted by Gastel
There is no inital hieght of the spout from which the water is flowing from.
12. 16 Mar '07 01:11
It states that the container is aligned with the bottom of the spout, indicating that there is height to the spout. Does the ten centimeter of height come after it reaches the top of the spout? This seems to me that there is no height to the spout.
13. 16 Mar '07 12:19
Wow, when people give you directions to their houses, you must have some real difficulty.

What do you think the question means?
14. 16 Mar '07 13:23
wouldn't you need to know the velocity of the water stream to determine where the peak of the ark would reside and how much of the 5cm of distance it has used when it peaks.
15. 16 Mar '07 14:18
Originally posted by joe shmo
wouldn't you need to know the velocity of the water stream to determine where the peak of the ark would reside and how much of the 5cm of distance it has used when it peaks.
Why would you need to know this? The information is described by geometric dimensions. If you want the velocity, then why not the thickness of the stream, the density of the water, the ambient pressure of the atmosphere?

The situation is already described, stop fighting it and accept the question.