Simple tapered bar - balance calculation ?

ddean

ddean

Joined
Mar 26, 2002
Messages
3,397
Does anyone know how to calculate the balance point
of a bar that tapers evenly along its length.

As a simple example, use:

Bar length = 1 meter

Maximum measures at one end
4 cm wide & 1 cm thick

Minimum measures at the other end
3 cm wide & .7 cm thick

I came up with a way to roughly estimate it,
a sort of manual addition of products based on
distance from one end and percentage of maximum dimension,
but I'd like something more equation-based.
 
I use my calculator and set it on a flat surface, next lay a pencil on the calculator and carfully balance the tapered steel across the pencil until both ends are off the surface the same distance. Then I simply use a magic marker at that spot. Works for me.
 
Takes a little calculus and a lot of algebra. You need to know the density of the steel,and the dimensions. Set up an equation relating the volume to the length (kind of tricky because of the taper, but it can be done) and multiply that by the density. That gives you the weight of a given length from the starting end. Set that equation equal to half of the total weight and solve for L the length, that gives you the balance point. At least theoretically. Nothing is perfect, so you'll end up having to fine tune it with a pencil or index finger in the end anyhow :)

If I get really REALLY bored later I'll set up the equation for the dimensions you listed. Its the weekend though, so chances are I won't :D
 
Yes, I am working on paper.

I want to develop a better --theoretical-- understanding of blade balance.

And I do understand that hands on is necessary.

At the same time I find the theory interesting and useful.

Keeps the brain cells active.
 
assume your bar is a truncated pyramid (rectangular base and top). the volume is ((area of base + area of top)/2) multiplied by height.

let x = altitude from base wherein the volume above is equal to volume below.

b = area at x

given: area of base = 4 sq cm
area of top = 2.1 sq cm
height = 100 cm

volume = 305 cu cm
half of which is 152.5 cu cm.

((4 + b)/2)*x = 152.5
2x + bx/2 = 152.5
bx/2 = 152.5 - 2x
b = (305 - 4x)/x

((b + 2.1)/2)*(100-x) = 152.5
50b - bx/2 + 105 - 1.05x = 152.5
-bx/2 -1.05x = 152.5 - 50b - 105
-bx/2 - 1.05x = 47.5 - 50b
x(1.05 - b/2) = 47.5 - 50b
x = (47.5 - 50b)/(1.05 - b/2)

substitute 'x' for the equation b = (305 - 4x)/x and solve for b

SHEESH I DON'T KNOW WHAT I'M DOING!!!!
 
Hey! It doesn't have to be that complicated. Drill a hole in the blade and put it on lawnmower blade balancer. :D

C Wilkins
 
Originally posted by hank_rearden .....SHEESH I DON'T KNOW WHAT I'M DOING!!!!
Click to expand...
Then I hope to aspire to that level.

Thanks. :D :D :D

I'll give it a try first chance.

Much appreciation
 
if it feels good do it.:D oh and for bal.I like the pencil way
if it takes more time to figure than to make, then I'm in the wrong business :) :)
 
The volume of a truncated pyramid is (h/3)*(A+sqr(A*A1)+A1), where h is the height, sqr is squareroot, A is the bottom area, A1 is the top area. You can form an equation to give you the CG point or you can do it by trial (which works surprisingly fast).

One other option is to use a 3D modelling program, draw the blade there, most programs can give you the CG. The good part is that you can "make" very complicated shapes where other approaches to CG calculations are very time consuming.

TLM
 
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