OT: Sword Blade Design Question

ddean

ddean

Joined
Mar 26, 2002
Messages
3,397
Sword Blade Design Question for the Engineers out there.

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.
 
If you have the slope of the sides and merely want to solve for the balance point:

I believe you would want to think of this as two polygons which have a height of (h-x) and (x). The balance point is the point at which the volumes of the two are equal. Express all the measures in terms (h-x) and (x) and solve for x. Hint: think of each polygon as consisting of one rectagular section and 2/4 triangular sections (this will make it easier to solve for volume).

n2s
 
N2? Why is gravity?
Click to expand...

That will work. :)

There is nothing better then a little physical testing to tell you where the balance point is. The theory is only useful during the design stage, while you are working on an electronic or paper draft, and have nothing tangible to simply test.

n2s
 
I am planning on starting to make knives in the next few months.
You can do what I am going to do.
Just make one and get better with experience, just have fun.
 
This doesn't actually answer your question... But it may be of interest to anybody interested in sword design dynamics. From "The Book of the Sword" by Richard Burton (1884)

"The late Mr Henry Wilkinson of London, a practical man of science, first proposed a formula for determining the centre of percussion without the tedious process of experimenting with each and every blade. His system was based on the properties of the pendulum. a light rod exactly 39.2 inches long, capped with a heavy leaden ball, swung to and fro upon a fixed centre, vibrates seconds or sixty times a minute in the latitude of London, and the three centres of percussion, of oscillation and of gravity are concentrated ithin the ball. If it were a mathematical pendulum - a rod without weight - these 3 points would lie precisely in the core of the ball, or 39.2 inches from the place of suspension. The balde, to be graduated, is suspended, tight-fastened at the point on which it would turn when making a cut, and is converted by swinging into a pendulum. As the length is shorter, so the oscillations are quicker; the blade makes 80 movements to sixty of the pendulum. a simple formula determines the length of such an eighty-vibrations pendulum to be 22 inches. This distance, measured from the point at which the blade was suspended, is marked on the back as the centre of percussion, where there is no jar, and where the most effective cut can be delivered."

(The Book Of The Sword, page 129)

Personally, I don't get it; but I always sucked at physics.

Presumably, if this technique can be used to find the CoP, it can be modified to find the centre of balance as well.

As for finding centre of gravity, trial & error strikes me as the easiest and quickest method. Pick up the item and move it around on the fulcrum until it balances...
 
Personally, I don't get it; but I always sucked at physics.
Click to expand...

The method mentioned by Burton simply takes into account the effects of gravity. On earth an object falls to the ground at a constant rate of 9.8m/S^2. A pendulum captures and uses this energy to swing so its motion is at a constant rate. The only thing that changes the number of swings that a pendulum makes within a given period of time is the distance that it has to travel. The further the distance, the longer it takes, and the smaller the number of swings per time frame.

If you suspend a sword at one end and swing it, it behaives like a pendulum. By measuring the number of swings per minute and finding the length of pendulum with the same number of swings, you can determine where the center of precussion is located. The CP will be at the same distance from the suspension point as the pendulum.

n2s
 
I'm just working on paper right now.
I'm a theoretical sort of guy.

Or more precisely, once I understand the theory I can learn
more quickly how to do. What to expect, look, feel for.

Nothing replaces hands on experience.
At the same time theory can help guide that experience.

Along the way, looking at theory, I'm looking to come up
with a way to estimate balance point and other factors
based on historical measurements of a style.
That will help me understand other things.
I'd like to figure out why the experts in two styles I've read on
(jian/gim & rapier) say that no one reproduces the 'real' thing.

I have a couple of blades (not the 'real' thing) to test as I
develop a theory of my own.

Even if I'm spinning my wheels, it's fun and keeps my mind from stagnating.
 
Thanks for explaining, n2s.

I still reckon picking the sword up and bashing on something with it is probably going to be quicker and easier.
 
Originally posted by nedesto .......i posted a formula for center of mass, NOT center of balance.
Click to expand...
Hmmm. I would like that too.

I would have thought they were the same, I'll learn something new.
 
okay then. to find center of mass:

formula for width as function of distance x along bar is:
W = .7 + .3x (more generally the formula is W = W0 + (W1-W0)x/L where W0 is the starting width, W1 is ending width, L is bar length)

similarly, height is H = 3 + x
area = H x W = .3x^2 + 1.6x + 2.1

integrate area to get volume = .1x^3 + .8x^2 + 2.1x
total volume (from 0 to 1) = .1 + .8 + 2.1 = 3.0

solve volume to find x when volume is 1/2 of total:
.1x^3 +.8x^2 +2.1x = 1.5 or .1x^3 +.8x^2 +2.1x -1.5 = 0

solve the cubic equation; the only real solution is .577
(easiest way is to plug coefficients into cubic solver ie http://www.1728.com/cubic.htm )

so center of mass in your example is .577 meters from skinny end.
1/2 the mass is on each side of this point. but not center of gravity
since the mass distribution is not the same on each side. it can be calculated but is more complicated.
 
....vibrates seconds or sixty times a minute in the latitude of London. ...

-----------------------------------------

Yeah, but how will that work here in the MidWest??? :p

Alan
 
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