



 
Robert, I labeled three points in the diagram you sent. Put a coordinate system on your diagram with the origin at O then the inside edge of the first gasket segment is part of the circle x^{2} + y^{2} = 28^{2} and the outside edge is part of the circle x^{2} + y^{2} = 30^{2}. The xcoordinate of P is 39.5980/2 and hence the ycoordinate of P is given by y^{2} = 28^{2}  (39.5980/2)^{2}. Likewise Q has xcoordinate 39.5980/2 and hence the ycoordinate of Q is given by y^{2} = 30^{2}  (39.5980/2)^{2}. The distance you need, the distance between A and B is the distance between P and Q which is
which I get to be 2.7399 (inches I presume). Your diagram gives this as 1.4142 + 1.3297 = 2.7439 so we agree to 3 decimal places. I hope this helps. Write back if you need anything further,  


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