- 28 Jan '17 17:30 / 4 edits

The short story is the metal does expand into the hole, but the radius of the hole grows larger at a greater rate.*Originally posted by robbebop***Can you explain why, because I was thinking the hole would become smaller, because the metal would enlarge in the hole?**

The long way to show this is "true" is to assume a model. This model has been verified experimentally over (for lack of a better word) "ordinary" ranges of temperature and temerature change.

The first assumption is the material has isotropic thermal expansion properties. This means that the material will expand at the same rate in ALL directions for a given change in temperature. The rate of this expansion is given by a property of the material known as the coefficient of linear expansion. The model says the change of length of a material for a given change in temperature is the product of the following three quantities: The coefficient of linear expansion of the material ( at or around the average temperature of the change);The initial length of the material, and the change in temperature the material undergoes.

The model I'm assuming works basically like this:

1) Calculate the intial length of materal from know geometric relationships ( the average circumference of the ring, which takes into account thickness of the ring)

2) Take the ring, split it and straighten it.

3) Subject the (now straigtened) ring to a change in temperature, letting it expand freely in all directions.

4) Take the expanded material at the elevated temp and reform it back into a perfect ring.

5) Re-calculate the radius of the ring at the elevated temp.

After you perform all the algebra in this model ( mathematical procedure outlined above). You will find that the New Radius of the ring is given by:

R = r + (α•δT)•r

In words that states the New Radius "R" will be the original radius "r" plus ( or minus depending on direction of temperature change) some porportion " (α•δT) " of the original radius "r".

So for positve temperature change (make it hotter) R > r , and for negative temperature change ( make it colder) R < r

Some of the more sophisticaed physicist might have some caveats (or reservations with the model) to add, but thats the best I could come up with. Hope that clears it up. However if you want to actually see the math that leads to the end result I'll type it up. - 28 Jan '17 17:37 / 5 edits

The whole circumference of the ring will increase; both the circumference of its most outer edge and that of the hole and by what is called in geometry the same scale factor ( see http://www.bbc.co.uk/bitesize/ks3/maths/shape_space/transformations2/revision/2/ ) although the absolute expansion, not to be confused with the scale factor expansion, of the hole would be a bit less than that of the outer edge of the ring; which is consistent with the thickness of the ring getting greater as a result of the thermal expansion.*Originally posted by robbebop***Can you explain why, because I was thinking the hole would become smaller, because the metal would enlarge in the hole?**

Heating, assuming the heat is evenly dispersed throughout the metal, will cause all of the metal to expand equally in all 3 dimensions so any metal object that is heated evenly will expand equally in all 3 dimensions so, whatever shape it is, it will stay the same shape except scale up in size by the same scale factor. The maths says a ring shape that stayed exactly the same exact kind of shape except that shape is enlarged by a scale factor would mean its hole will increase by the same scale factor. - 30 Jan '17 16:22

Wow, never thought about it that way, a very good illustation of the problem. Thank you very much!*Originally posted by twhitehead***To prove it to yourself without an actual metal ring and an oven, take a bunch of coins of two different sizes. Put the small coins in a ring and the same number of large coins in a ring. Which ring has a bigger hole? Now just imagine the coins are atoms.**