MetalworkingFun Forum

How do you determine the safe capacity for v-blocks?

I see some being sold with the capacity listed and others without.

No idea :)

I needed to straighten a 2 1/4 inch diam bale spike for a neighbour in my 60 ton press, and foolishly used my largest Vee blocks - only Chinese and nothing too precise. I was aware that it was a silly thing to do as they are (were!) cast iron. Fortunately my press has an umbilical lead for the Up / Down buttons so could stand to one side. Made a fair old 'crack' when they crumbled :)

I should have been clearer in my question. I meant the diameter of the work piece.

This is just a SWAG but I think that it would be the largest diameter that rests on the Vs since that's the precision surface. It is a good question though.

Ed

Pythagorean theorem is your friend, assuming you have the dimensions of the Vee.

Measure the width across the top (widest part) of the Vee and draw a simple sketch. That distance (perhaps a little less to assure the round is sitting on 45º planes), divide by 2. Now use that number, square it, multiply X2, square root of the result, multiply by 2, that's your maximum. And it may not fit into a slide-in clamp. If the Vee blocks have that, measure across it's full inside radius and subtract a tiny bit.

I'll have to put my thinking cap on Ken, as that is all Greek to me!

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

Thanks Ken - I understand it. I was having a little joke with the "its all Greek to me", given that Pythagoras was a Greek mathematician... I'll measure them up next time I'm in the shop.

“V” blocks usually come in pairs and some come with clamps. The internal width of the clamp will determine the maximum diameter of the bar.

Removing the clamps will allow for a larger diameter bar, how much larger? That just comes down to “Good Mechanical Engineering Practices”

What is not considered good practice is using “edges” (sometimes referred to as corners). Using edges is bad – just not an accurate surface. This is exactly what Ken said using some “Greek maths”

Please note the abbreviation of the word Mathematics is ‘Maths” not Math.

Having said all that there is nothing to stop one placing a large diameter bar on a small “V” it just means the bar will be sitting on the “V” block edges and therefore not considered “good mechanical engineering practice” still doesn’t mean it cannot be done.

Just using the eye will say if the bar is stupidly too large for the “V” block.


DaveH

"Please note the abbreviation of the word Mathematics is ‘Maths” not Math. Big Grin Big Grin Big Grin"

It's these poor old 'Mericans - never did take a dictionary with them on the Mayflower