Croquet mallet head

You have a volume of 50 x 50 x 280 mm  =  700000 cubic mm

The mass is                                                    1850 gm

 

 

You want to reduce the mass by ,drilling holes,to 1333 gm.

So you need to reduce the volume to (1333/1850) x 700000 cubic mm

which is a volume of                                                   504378.378  cubic mm

 

Subtract  the new volume from the old one to see how much you need to cut away

 

that is    700000  - 504378.378  = 195621.621 cubic mm

You didn't say whether you would drill lengthways or sideways thru the block,but either way,the volume of each hole you drill is just  (pi)(r^2)   x  length of hole,where r is the radius of the hole,that is the radius of your drill. (Make sure it's the radius,and not the diameter,or you will be removing more.)

 

Happy drilling!

Hi Robert, 

There are probably elements of your question that were not explained - like how long can a drilled hole be but this is my answer based on the information given and my understanig of the problem :)

 

Hello

Please excuse the possibly simple arithmetic question.

A friend has asked me to make him a croquet mallet head and has supplied me with an aluminium block measuring 280mm X 50mm X 50mm.  

Volume = 280*50*50 = 280*50*50 = 700,000mm^3

This block weighs 1850gm.

 

I need to reduce the weight to 1333gm.

Need to make holes to remove 1850-1333=  517grams

517/1850*700000 = 195621.6216216216216216

Need to remove 195,622mm^3

 

 

To do this I intend to drill a number of holes through the block from side to side.

 

I must drill an equal number of holes on each end of the mallet so that it is balanced.

 

I am comfortable working with with wood but aluminium is new to me and I don't want to muck it up.

 

I have a set of drills of the size (Murphy was an optimist) 9/16", 5/8', 11/16", 3/4", 13/16", 7/8", 15/16" and 1".

I also have a 2 full sets of both metric and inch drills up to 13mm and 1/2" respectively

 

The biggest diameter that you mentioned was 13mm So I will use that one.

The dimensions given are in mm so I am not even going to consider the imperial bits.

 

 

2pi*6.5^2*x=195,622

x=195622/(2*pi*6.5^2) = 736.9043379295325296

The block is only 280mm deep so

736.9/140 = 5.2635714285714286

122.8166666666666667 = 122.8166666666666667

6 holes on each side, each one has 13mm diameter and 122.82cm length

 

Check

6*2*pi*6.5^2*122.82 = 195626mm^3

195626-195622 = 4

So it is out by 4mm^3 which is negligable.