Game Development Reference
In-Depth Information
Figure 23-1. Pressure differential
It is clear that the water pressure acting on the sphere is much larger than the air pressure
we trapped inside before sinking it. Also, note that pressure always acts normal to the
surface. If you happen to apply a force to the vertex of an object, you'll have trouble
modeling the right effect because a vertex does not have a well-defined normal. We can
overcome this only by applying pressure to the faces of polygons or by averaging the
direction of the pressure on either side of the vertex. Returning to our example, the net
pressure differential on the steel ball is:
P(water) − P(air) = 10,949,100 N/m 2 − 101,000 N/m 2 =
10,848,100 N/m 2
This is the pressure you would feel if 1,870 elephants were standing on a 1-square-meter
plate on top of you. If our steel ball had walls that were too thin to withstand this pressure
differential, it would implode. To put this all into perspective, a steel ball in space would
have a pressure differential of only 1 atmosphere pushing out. It is thus much harder to
design a structure to go to the bottom of the sea than it is to go to the moon. Indeed,
more men have stepped foot on the lunar surface than have visited the bottom of the
sea.
 
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