alext87 Posted December 4, 2005 Share Posted December 4, 2005 I am doing some analysis of results from an impact crater caused by steel balls. I am measuring the diameter of the impact crater and need to link it to the volume of the crater using the volume of resolution. Can anybody help I have a formula that I derived it is correct? Volume of moved by impact = (2/3) π r3 – (1/12) π(4r2-D2)3/2 where r is the radius of the steel ball and D is the diameter of the crater. This i think is linking the diameter of impact crater to the volume moved... I also have two other calculus problems: I need to find out the surface area of the crater in terms of the diameter of the impact crater can anybody help??? Then I need the rate of change of surface area with respect to diameter and then to depth??? These problems are beyond my mathematical skills... Link to comment Share on other sites More sharing options...
insane_alien Posted December 4, 2005 Share Posted December 4, 2005 don't you also need the depth of the crater to work out its volume? Link to comment Share on other sites More sharing options...
alext87 Posted December 4, 2005 Author Share Posted December 4, 2005 As (diameter of crater/2)^2 = (radius of the steel ball)^2 + Radius of steelball - depth of crater)^2 which is just the equation of a circle. There is a definite relationship so that if you know diameter of the crater you also know the depth. Link to comment Share on other sites More sharing options...
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