High School A tilted container with water inside - how to calculate the distance A?

[DoyLin]

TL;DR

Geometry

We are trying to control a container with water inside. The container has two phases of operation. In one phase, the container is tilted. In this tilted state, the distance from the observation point (red dot) to the surface of the water (distance B) is known from a measurement device. However, we actually would like to know this distance when the container is in a flat position (distance A).


In the control system, the dimension of the container is known (C and E). The tilt angle (alpha) and the distance B are known from measurement devices. The position of the observation point is known (D).

The container cannot positioned in a flat position. But we want to know the distance A (distance when the container were laid flat).


Could anyone suggest how to calculate A ?


Thank you very much, Doy

 

[Nugatory]

You can calculate the volume of the liquid from the dimensions of the box (although you are missing one - a box has length, width, and height) and the $B$ and $\theta$ measurements. Calculating $A$ from that volume is easy.

 

[Lnewqban]

Welcome, @DoyLin !

Consider that the point of the surface of the water that is located at C/2 remains at a distance A from the top of the container for any tilt angle.

 

[Chestermiller]

Welcome, @DoyLin !

Consider that the point of the surface of the water that is located at C/2 remains at a distance A from the top of the container for any tilt angle.

Not if $\tan{\alpha}\gt E/C$

 

[DoyLin]

You can calculate the volume of the liquid from the dimensions of the box (although you are missing one - a box has length, width, and height) and the $B$ and $\theta$ measurements. Calculating $A$ from that volume is easy.

Hi, thanks a lot for your feedback. However, we don't know the amount of water inside the container. The container is not fully filled with water. What we can measured is only the angle alpha and the distance from the observation point to the surface of water.

 

[Nugatory]

However, we don't know the amount of water inside the container.

Of course you don't know the volume of water in the container - if you did the problem would be trivial.
However, you can calculate that volume from the information you do have: the tilt angle, the distances $B$ and $D$, and the dimensions of the box.

I won't do the math for you because this forum is about helping people find the answer for themselves instead of handing them out, but just knowing that you can calculate the volume should be enough to get you on the right track.

 

[Chestermiller]

In Fig. 2, draw a line at an angle of $\alpha$ to the horizontal at an elevation of E-B above the base at x =D. The equation for this straight line (using the lower left corner as the origin) is $$y=x\tan{\alpha}+b$$where b is the intercept at the left, with $$b=(E-B)-D\tan{\alpha}$$

 

[bob012345]

Can one choose the observation point D or is it fixed?

 

[Chestermiller]

Can one choose the observation point D or is it fixed?

Apparently not.

 

[bob012345]

Apparently not.

In your solution for $b$, did you assume $tan(α)<E/C$?

 

[Chestermiller]

In your solution for $b$, did you assume $tan(α)<E/C$?

Sure. I was trying to keep it simple.

 

[bob012345]

Hi, thanks a lot for your feedback. However, we don't know the amount of water inside the container. The container is not fully filled with water. What we can measured is only the angle alpha and the distance from the observation point to the surface of water.

You can get $A$ from your measurements and what @Chestermiller showed in post #7.