As a matter of curiosity:
If one cubic yard of solid limestone is crushed, what volume of 3/4-inch clear stone will be produced?
I ask because the clear stone will have gaps between the granules. So it is my assumption that the volume of the clear stone will be greater than the source. The screenings/fines would be discarded.
This question is unanswerable. Different deposits of limestone can have different properties. The difference between in-situ volume (measured in bank cubic yards or meters) and the loose volume (measured in loose cubic yards or meters) following excavation & crushing is due to the swell factor. This accounts for the voids introduced in the loose material but also an expansion of the individual stones/particles because they are no long subject to confining stress, as they were when they were in-situ in the original deposit.
Two sources that I found online state the swelling factor can be 63 percent, or range between 75 and 80 percent.
As to the proportion of stones that are 3/4 of an inch in size after crushing, this will be affected by the properties of the limestone, for a particular deposit, how it was excavated and how it was crushed (jaw or gyratory crushers).
The same way that different soils will produce different particle size distribution curves, so will crushed stone, including crushed limestone. The settings for the crusher used will factor into this as well.
Just as a rough estimate.
I assume cutting one cubic yard of limestone clean with no sawdust. Further, assume that this 3.4-inch cubic yard of limestone is poured to assume a heap of cone shape resting at the angle of repose.
The typical void ratio for crushed gravel which has some smaller aggregate in the mix, so packs better is 40%-50%.
So assuming 60% packing efficiency, the volume of the crushed limestone will be.
$$V_{crushed}=\frac{V_{total}}{0.6}=1.66V_{total}$$
