Let me know what you are having trouble figuring out.
Here's an example:
Suppose the width of the "vee" at it's top measures 2.125" and we want to know what is the biggest diameter we can safely, securely hold.
Start with a little "safety factor", say we want the widest contact points to be at 2.075" wide. Divide that by two, equals 1.0375. Now "square" that figure, so 1.0375 x 1.0375 = 1.07640625. We know that for 45º right triangles the sides have to be equal, so in true Pythagorean Theorum fashion where we are looking for the length of the hypotenuse you would add the side adjacent squared to the side opposite squared. In this 45º case, 1.07640625 + 1.07640625 = 2.1528125. Now the square root of THAT number is the length of the hypotenuse. More specifically its the distance from the contact point of the round on the 45º surface to the center of the round, 1.4672465709620861131317520513676. That hypotenuse is the radius of the round, so double that is your maximum DIAMETER, 2.934493.....bla bla bla
Here's an example:
Suppose the width of the "vee" at it's top measures 2.125" and we want to know what is the biggest diameter we can safely, securely hold.
Start with a little "safety factor", say we want the widest contact points to be at 2.075" wide. Divide that by two, equals 1.0375. Now "square" that figure, so 1.0375 x 1.0375 = 1.07640625. We know that for 45º right triangles the sides have to be equal, so in true Pythagorean Theorum fashion where we are looking for the length of the hypotenuse you would add the side adjacent squared to the side opposite squared. In this 45º case, 1.07640625 + 1.07640625 = 2.1528125. Now the square root of THAT number is the length of the hypotenuse. More specifically its the distance from the contact point of the round on the 45º surface to the center of the round, 1.4672465709620861131317520513676. That hypotenuse is the radius of the round, so double that is your maximum DIAMETER, 2.934493.....bla bla bla