Maths conundrum

[acjy1985]

We have been doing some tests that give a 2d picture of a sphere(so a circle). Using this circle it is possible to deduce the volume of the sphere using 4/3 x pi x r3.

Our sphere volumes can differ by +/- 10%. So how do we calculate what the 2d circles can differ by to result in this 10% change in volume?

 

[NissanSkylineN1]

What's the reason behind the +/- 10%?

 

[acjy1985]

The results can mean a difference in volume of +/- 10% to account for error.

 

[Pezzarinho17]

4/3 x pi x r3 = Standard Volume (SV)
SV x 0.9 = Lower Volume (LV)
SV x 1.1 = Upper Volume (UV)

Then use:
r= Cube Root((3V)/(4pi)
Replace V with your values for LV and UV (so you do calculation twice) and you will get your pair of radii required.

EDIT: Just read question again, I think this is what you mean?

EDIT 2: Or I would just cheat and model the sphere in 3D CAD

 

[acjy1985]

Ok, that's great. Thank you for this. Is there a way to work out if you know the volume has changed by 'x' amount, then the radius must have changed by 'y' amount?

 

[Pezzarinho17]

Ok, that's great. Thank you for this. Is there a way to work out if you know the volume has changed by 'x' amount, then the radius must have changed by 'y' amount?

Use this, you just need to change your volume and you will get the new radius value.
r= Cube Root((3V)/(4pi))

 

[acjy1985]

Can you believe I did A level physics?

Anyway, thanks so much for your help.

 

[Pezzarinho17]

Can you believe I did A level physics?

Same, but the A level Maths is more useful here!

Anyway, thanks so much for your help.

👍

 

[BobK]

For the volume to be 10% greater, the radius would have to be the cube root of 10% greater (ie, cube root of 1.10). The projected circle would have the same radius, so its increase in area would be proportional to this cube root, squared. Or roughly 6.6% greater.