Partial volume of a hexagon

K

Kobus

New member
Joined
Nov 27, 2019
Messages
1
Good day, I need assistance to calculate the volume of water in hexagon and pentagon shaped containers lying horizontally. The volume of water would usually be under the half-way mark.
 
I think that the volume of water would depend on how much below the the half way mark the water is? What do you think? What have you tried? Do you know the volume formula for a hexagon and a pentagon? Do you know the area of the base of these figures? You need to give us something so we can help you. Please read the forum's guideline so you know how to receive help.
 
Assuming the containers you have in mind are prisms with their horizontal base being a hexagon or pentagon, with vertical flat sides, the volume will just be the area of the base times the height (depth of water).

You haven't told us what dimensions you know about either container. If you happen to know the base area, then I've just told you all you need to know. If you know the volume when full, then the volume at a certain depth will be proportional to the depth. Otherwise, you'll have to tell us the details. In particular, are these regular polygons (all sides and angles equal), or not?

And if the base is not horizontal, we'll need lots of information!
 
You must log in or register to reply here.
Top