this post was submitted on 30 Aug 2026
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[–] NaibofTabr@infosec.pub 156 points 3 days ago* (last edited 3 days ago) (2 children)

On further inspection, it was found that the mathematician had assumed an ideal sphere of constant radius, while the real object was in fact oblate. The mathematician had neglected to take any measurements, leading to an incorrect value. The physicist's answer was correct, but only at sea level at nominal temperature.

The engineer is currently using calipers to measure the diameter of the object at many different angles in order to build a CAD model. He said we could expect results next year.

[–] tyler@programming.dev 77 points 3 days ago (3 children)

The physicist’s answer would be correct anywhere, no matter the temperature or density of the fluid.

[–] montechristo@feddit.org 32 points 3 days ago

Supposing that the measurements with and without the sphere are taken under identical circumstances. But you deserve the upvote either way.

[–] OwOarchist@pawb.social 20 points 3 days ago* (last edited 3 days ago)

Interestingly, the physicist's answer would always be correct, but not always consistent.

If the sphere is compressible, then air pressure will slightly affect its volume. And unless the thermal expansion rate of the sphere is zero, the amount of fluid displaced will also vary with temperature.

However, even if the physicist gets different answers in different environments, it's always still a correct answer, because they correctly measured the volume of the sphere at that time.

[–] untorquer@quokk.au 14 points 3 days ago* (last edited 3 days ago) (3 children)

Assuming the time between submersion and observation is greater than zero, increasing temperatures above boiling would give increasing error.

Conversely the physicist may get other error at/near freezing.

And yet another option is the fluid being above the Ball's melting and/or boiling point, assuming changes in density with phase change.

[–] tyler@programming.dev 2 points 2 days ago (1 children)

This is like saying “the physicist’s answer would have errors because they measured the ball and then stuck it in an oven. The volume of the ball in the oven is different than what they measured.”

[–] untorquer@quokk.au 0 points 1 day ago (1 children)

No it's more like the physicist tried to use boiling water to measure it

[–] tyler@programming.dev 1 points 1 day ago (1 children)

Using boiling water is still fine, the measurement will work, your comment literally only is a problem if they don’t measure in time. So yes, it’s exactly like sticking it in an oven and not measuring at the same time

[–] untorquer@quokk.au 1 points 1 day ago

Fair, if you do account for the change in water volume over time then you could maintain accuracy, but also good luck reading the meniscus.

[–] cloudshouter@lemmy.zip 14 points 3 days ago (1 children)

They just said the ball was red, they didn’t say why…

[–] untorquer@quokk.au 1 points 1 day ago

Responded to inline comment saying any temp would be fine

[–] Aqivex@fedinsfw.app 11 points 3 days ago (1 children)

All of which the physicist can trivially account for, control and variable test around, should the requested question be updated to require it.

The requested question did not include "In all possible scenarios", thus determining the volume of the sphere via physics calculation at the most average of conditions, is the most correct valid solution, absent further requirement parameters.

[–] untorquer@quokk.au 7 points 3 days ago* (last edited 3 days ago) (1 children)

You're right in terms of grading a student's homework or whatever.

In research/publication/career these assumptions must be stated unless given.

But I was responding to:

The physicist’s answer would be correct anywhere, no matter the temperature or density of the fluid.

Which includes non-standard conditions.

[–] Aqivex@fedinsfw.app 2 points 1 day ago

But I was responding to: The physicist’s answer would be correct anywhere, no matter the temperature or density of the fluid. Which includes non-standard conditions.

[–] BartyDeCanter@piefed.social 5 points 3 days ago

Anywhere with gravity sufficient to overcome the surface tension of the water from forming around the object or clinging to the container.