Asked by bob
Hi! Can anyone give me some tips on how i would determine the shape of a vessel from a graph of volume vs time?
"a set volume of water is poured steadily into a vessel, the height at set volume intervals is recorded." "determine a possible vessel shape based on this data"
Volume Height
1 1
2 5
3 12
4 16
5 21
6 24
7 25
I have made a graph of this, and then used gradients to find a possible shape by using the amount of water increases at each step..
any ideas? thanks
"a set volume of water is poured steadily into a vessel, the height at set volume intervals is recorded." "determine a possible vessel shape based on this data"
Volume Height
1 1
2 5
3 12
4 16
5 21
6 24
7 25
I have made a graph of this, and then used gradients to find a possible shape by using the amount of water increases at each step..
any ideas? thanks
Answers
Answered by
bobpursley
The maximum gradient occurs at the bottom, so this suggests a bowl shape to me.
Answered by
Damon
The surface area of the water times change in height = change in volume approximately
So, we can start finding the cross sectional area as a function of height.
for example assume the area is 0 at 0 height
then at height one the volume is one
so the average area between 0 and one is
A = 1/1 = 1
now between volume of 1 and volume of 2 the average area is
1/(5-1) = 1/4
between volume of 2 and volume of 3 we have
1/(12-5) = 1/7
between volume of 3 and volume of 4 it is
1/(16-12) = 1/4
etc
if that area is square, then the sides are sqrt area
if the areas are circles, the radius = sqrt(area/pi)
So, we can start finding the cross sectional area as a function of height.
for example assume the area is 0 at 0 height
then at height one the volume is one
so the average area between 0 and one is
A = 1/1 = 1
now between volume of 1 and volume of 2 the average area is
1/(5-1) = 1/4
between volume of 2 and volume of 3 we have
1/(12-5) = 1/7
between volume of 3 and volume of 4 it is
1/(16-12) = 1/4
etc
if that area is square, then the sides are sqrt area
if the areas are circles, the radius = sqrt(area/pi)
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