Author 
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Thomas Riley

Posted: Thu 15 Sep, 2016 12:02 pm Post subject: Sword Form and Thought Chart creations 


Hello,
I've been doing my own research, after reading the book "The Sword: Form and Thought" and working on applying the concepts to my own collection.
In it are graphs which describe the sword "at a glance":
I understand how all of these things have been measured and graphed, except for 2 aspects: The Oval and the Cones, which represents the agility of the sword. I understand the Oval represents linear transition agility, and the cones are some form of rotational agility.
I've read over George Turners "Dynamics of HandHeld Weapons" to dive into the math for a lot of this, but in the end have not been able to crack the nut of these 2 aspects.
Perhaps someone here can help me with it. How exactly are they determining how to chart these things, specifically the magnitude of both the oval, and the angle for the cone? Dum Spiro Spero, Dum Spiro Scio 



J. Nicolaysen

Posted: Thu 15 Sep, 2016 4:36 pm Post subject: 


I am probably wrong, but I assumed the oval and cone correlated as how far you'd move the hilt (oval) to get the tip and edge to the side of the cone. So basically, the smaller the oval and wider the cone, the more agile sword you have, minimal effort with maximum effect. I'm looking at the red dots to see this. If you mark a dot on the hilt traveling to a point on the oval, you can measure the dot on the cone...
But I am very interested to hear an expert's thoughts on this, it's just my quick understanding from the catalogue.
edited to replace "Dots" instead of triangles, sheesh. Talorgan Drust on FB 



Bram Verbeek

Posted: Thu 15 Sep, 2016 11:46 pm Post subject: 


I think it would not be hard to do.
Rotational: Set sword on a turning table with constant known resistance, centered on the center of mass. Give a constant known impulse at the handle, measure the angle the sword travels.
Translational: Set the sword at a slide with a constant known resistance, give the sword a constant known impulse at the handle and measure how far the sword travels.
The hard part is to gather why the ovals are not round. 



Thomas Riley

Posted: Fri 16 Sep, 2016 4:16 am Post subject: 


Bram Verbeek wrote: 
The hard part is to gather why the ovals are not round. 
So I think it is an oval because of the effective mass at that part of the handle, that is the only thing I can think is causing that. I'm assuming that this is mathematically derived, and that these measurements are not based on actual tests, mostly because this was performed on very old originals, and I would assume they only really had access to the basic specs; weight, length, etc..
Again I just don't understand how they came up with the sizes. There does seem to be a pretty direct correlation from weight to size of ovals and cones; the lighter is it, the larger the cone and oval. The cone is also based on length of blade, so I assume there is some correlation between the weight and PoB, to get themselves an angle between 0 and 90 degrees, and then built it based on that. It's just that it is never explained in the book, so I only really get a graph that cannot be replicated. Dum Spiro Spero, Dum Spiro Scio 



Vincent Le Chevalier

Posted: Sun 18 Sep, 2016 12:48 am Post subject: 


Hello all,
I happen to be responsible for the construction of these graphs
The goal in that book was to get a visual impression of various aspects of the sword's balance, that could otherwise only be perceived sword in hand. It's not really mature as a comparison system; this is still a work in progress for me and Peter. We also had a fairly limited space and had to explain all the basic concepts first, and this is why the explanations are not sufficient on these diagrams.
The oval and the cone are meant to represent the potential for acceleration of the weapon when using several modes of motion. Mass is the primary factor in their size, and mass distribution influences the relative shapes. The bigger the mass, the smaller the shapes become, showing the difficulty in accelerating a heavy weapon. The oval represents how the sword accelerates in different directions when subject to a force applied at the cross. So if you push along the axis of the weapon, you meet the whole mass. If instead you push perpendicularly, you only meet part of it: the effective mass at the cross. That is why these are ovals and not circles. The cone represents how the sword accelerates when moved in rotation around the cross. It also takes into account leverage; these diagrams are computed assuming constant force at the end of the handle, basically. Otherwise twohanded swords would have a much smaller cone.
So to compute them you need:
M, the total mass of the weapon
H, the location of the cross
G, the position of the center of gravity
R and P, a pair of associated pivot points
L, the location of the handle end (pretty much the aft location on the handle), where leverage is applied
and then you compute:
J = GR*PG + GH^2
Width of the oval = ko/M
Height of the oval = ko/M * J / (GR*PG) [EDIT: made a mistake!]
Angle of the cone = kc/M * HL / J
The coefficients of proportionnality ko and kc were picked to give a proper spread. You should probably pick your own for your collection (depends on the units you measure in as well)...
Hopefully a much more complete publication about sword balance will follow in the future!
Hope this helps, 
Vincent
Ensis Sub Caelo
Last edited by Vincent Le Chevalier on Wed 21 Sep, 2016 8:55 am; edited 1 time in total 



Thomas Riley

Posted: Sun 18 Sep, 2016 5:03 am Post subject: 


Vincent Le Chevalier wrote:  Hello all,
I happen to be responsible for the construction of these graphs

Well first off; Thank you! I find these graphs quite fascinating and useful.
I ended up reverse engineering the method and resulting in different formulas, which seem to align with what you provided. The main issue was trying to replicate your coefficients of proportionality, or in my case a force constant of .2 (force and torque). I used standard rotational physics formulas and the George L Turner research paper.
I ended up using a sword in my collection that had extremely similar physical properties (as best I could match) to a sword in the book, and then I was able to use information such the the pendulum period and to figure out some of the underlying math.
I wish I had seen your reply yesterday when I started this number crunching effort, but my efforts were not in vain, and from it I created a quite lovely graph for my Albion Liechtenauer:
I will go back and check my results against your approach, but I'm pretty confident this is an accurate model, just from a different approach.
Thank you! Dum Spiro Spero, Dum Spiro Scio 



Vincent Le Chevalier

Posted: Sun 18 Sep, 2016 6:46 am Post subject: 


Looking good Thomas!
The whole thing hinges on accurate measurement of pivot points, and this is something we spent quite a lot of time exploring. Here I'm a bit surprised with the location of the aft and forward pivot that you have found; on this type of sword I'd expect the forward pivot to be in between the blade node and the tip, and the aft pivot further from the center of gravity. I'm not sure how you measured them, I guess some form of waggle test?
Things to look for with the waggle test:
 It is easier to do with the blade down
 As little torque as possible should be applied to the sword. I hold the sword just between two fingers
 The frequency of the motion must be as high as possible. If you move too slowly the fixed point tends to migrate back to the hilt
 Spotting the fixed point can be difficult on smooth blades. I use a hair tie on the blade that I move up and down until I see it fixed. Also makes it easier to measure, as you can move the sword around without losing the point.
The values on the curve are computed just like the hilt agility (inverse), so at point X:
value = M * GR*PG/(GR*PG + GX^2)
That should allow you to build a curve and determine the mass and fraction at node, consistent with the pivot points.
Regards, 
Vincent
Ensis Sub Caelo
Last edited by Vincent Le Chevalier on Wed 21 Sep, 2016 9:53 am; edited 1 time in total 



Thomas Riley

Posted: Mon 19 Sep, 2016 3:58 am Post subject: 


Thanks! I was surprises by the location of the pivot node and vibration node locations, but I measured it multiple times... I like your approach of using a band of sorts to assist. I used a sticky note.
Anyway I'll incorporate your formulas in the future to make all this easier on myself.
Thanks Again. Dum Spiro Spero, Dum Spiro Scio 



Thomas Riley

Posted: Mon 19 Sep, 2016 7:58 am Post subject: 


Quick question, regarding the formula: J = GR*PG + GH^2
In this case GR represent the distance between the Center of Gravity and the pivot point #1, PG represents the distance between the Center of Gravity and the pivot point #2, GH represents the distance between the cross and the Center of gravity.
So when G = .375m, R = 0.77m, P=0.26m, H=.285m, L=0.06m
then
GR = 0.395m, PG = 0.115m, GH=0.09m, HL=0.225m and J = 0.053525
Am I doing this correctly? Dum Spiro Spero, Dum Spiro Scio 



Vincent Le Chevalier

Posted: Mon 19 Sep, 2016 9:50 am Post subject: 


Seems right yes!
Regards, 
Vincent
Ensis Sub Caelo 



T. Kew

Posted: Mon 19 Sep, 2016 10:11 am Post subject: 


I agree with Vincent that the pivot points you've presented are dubious.
Maarten Kamphuis (of mBlades) has presented some results about the behaviour of swords, and how they're related to mass distribution, pivot points, and so on.
The most relevant one for this discussion is the following:
a * m * b = k
Where a is the distance from a pivot point in the hilt to the centre of mass, m is the mass of the weapon, b is the distance from the centre of mass to the matching pivot point in the blade, and k is a constant for the sword (determined by its overall mass distribution).
As a consequence of this, if you double the distance from your hilt pivot point to the centre of mass, you halve the corresponding distance in the blade.
This relationship doesn't seem to hold true for the set of data you've diagrammed on your sword.
(This also has a geometric expression  if you draw a circle through a rear pivot point and its corresponding forward one, the point at which that circle intersects a perpendicular drawn from the centre of mass is constant for all pairs of pivot points. That also doesn't seem to be true, but I've only poked the diagram a bit with a set square) Instructor and scholar, Cambridge HEMA 



Vincent Le Chevalier

Posted: Mon 19 Sep, 2016 10:19 am Post subject: 


Indeed that is the manipulation I did too. It is pretty useful to check several pairs to make sure there isn't a faulty measurement somewhere, and realize that no measurement is perfect...
(See my article about the geometrical interpretation too!)
Pivot points should also correlate with pendulum tests, if that is how you computed the curve. The length of the pendulum should be the same as the distance between the center of rotation and its pivot point.
Regards, 
Vincent
Ensis Sub Caelo 



Thomas Riley

Posted: Mon 19 Sep, 2016 10:22 am Post subject: 


I'll be revisiting the Liechtenauer this week brand new and fresh, remeasure everything, and use some of the methods described in this conversation. Dum Spiro Spero, Dum Spiro Scio 



Nathan Robinson
myArmoury Admin




Thomas Riley

Posted: Mon 19 Sep, 2016 12:51 pm Post subject: 


I have, that is me.
I will be updating my approach as I process and understand what was covered in this thread. This thread picked up after I had made that video. Dum Spiro Spero, Dum Spiro Scio 



Nathan Robinson
myArmoury Admin




Thomas Riley

Posted: Tue 20 Sep, 2016 10:53 am Post subject: 


Preview of the new model for the Albion Liechtenauer:
You can see it is very different from my first version, and I feel it is much more accurate. Especially the vertical hilt agility. This is using only the formulas provided by Vincent.
With some help from a YouTube Commenter I can now provide this method for quickly generating the graph information on GeoGebra: https://ggbm.at/yqYH2F2h
Later I will build the final vector graphic for all of this, and will post the final version as an FYI for anyone following this thread.
Thank you all! This is really awesome. Dum Spiro Spero, Dum Spiro Scio 



Thomas Riley

Posted: Tue 20 Sep, 2016 4:27 pm Post subject: 


Final version:
Thank you so much for all the help! Dum Spiro Spero, Dum Spiro Scio 



Nathan Robinson
myArmoury Admin




Thomas Riley

Posted: Tue 20 Sep, 2016 6:29 pm Post subject: 


As I look through the book, it strikes me that almost none of the swords have this kind of ratio for the linear agility, why am I getting such tall ovals in the sword I am trying this on?
was the constant "ko" different for the X and the Y axis? Dum Spiro Spero, Dum Spiro Scio 



