no need for japanese mambo jambo

Somewhere I read that all the info you need to become a good woodworker was free. I totally agree with that. And I think it's morally reprehensible to put knowledge behind paywalls of any kind. One doesn't have money because one is smart or worked hard. It's mostly related to the special skill of choosing one's own parents. Born white and with money in a first world country sets you on top of a few million people, if not billions. Make it even 20cents dearer for them to know what you know seems like a crime to me. Mostly when you, your parents, and your grandparents have profited of the exploitation of those people be it by slavery, economic dependence or simply by producing far more of your share of pollution to have the life you have and don't give a shit.

So when my friend Gabe speaks in japanese mambo jambo about triangles I go back to my old friend pitagoras and start from there to splay legs. Pitagoras theorem is free, like the concept of wheels and diesel engine. Google's algorithms are not. Neither several life saving medicaments or technologies. Without been boring about the whole story of the concept of copyright, it stems from the same place as capitalism, colonialism and the like: england 1600s. And as every reactionary idea, it attempts to present as natural what was once imposed.

That's why there is a little sign at the end of my blog that says something like all this crap is creative commons. Ya know, it's supposed to help the common good.

So today I used the spare leg for one side, and then made another one for the remaining one.

The first leg is "just" 3mm undersized, well, in fact the mortice is oversized. The others are quite decent. We needed to have dinner tonight so the bench was moved down before I could finish plane it. Once I put shellac on it I show some close ups.

It's much easier than what I feared, and you can solve this with simple geometry and pitagoras theorem. In the case of this 50x50mm legs I needed to cut 3mm from each side to make the diamond shape.

A few tips: cut the bottom of the legs at the end. If you already cut one bottom, you can use it as a paring block. Make a paring block if you haven't cut a leg's bottom. And not only use it for paring but also for holding the chisel at an angle when you hammer it.

 The bench sounds like crazy when we sit. Well, mostly when I sit, it holds julia without complain. The dog sleeps under it. I hope it doesn't break.

mirrors

Gabe said something about 2 legs being the mirror of the other. I thought, and it made sense to me, that since the mortices where there those go in one way only, so it needs to be so. But in the case without mortises not anymore.

So I went ahead and since I'm skipping the stretchers just marked and cut the 4 legs in the same way.

Now I have two spare legs or a frankestein extension work for tomorrow:

Or maybe our floor is so uneven that it will work anyway. I had anticipated that something was going to go wrong, so I had 1 spare leg. But I never thought it could go so wrong. Well, live and learn. 

It's looking nice —imho — though. Way better than the paint buckets that it used to have as legs. Tomorrow I plan on finishing the mortises and surface planing the top.

Thursday I go buying wood and want to make one or two more. By the 5th mortice I think I will get the hang of it.

legs

Mara macho. Male mahogany that is. Or so called in Peru or wherever this bastard comes. I have problems sanding Rosewood but I'm getting allergy planing this boy. Real annoying wood and the grain is a bastard.

This week I'm going to get more spanish cedar, which actually comes from Bolivia and not Spain.

The legs, as you can see are diagonalised. Since the dog sleeps under the bench where these girls go I won't put stretchers to it, so it's a mix of staked and japanese bench. With spanish cedar on top. I may re make the legs one day since I'm not so happy with the grain of these one.

My little bevel gauge is no the best for marking out, you want something long so you can transfer the slope from the sashigane to the gauge easily. I want one like this: http://granitemountainwoodcraft.com/2016/04/10/splay-leg-layout/

You see Gabe, I took the challenge.

frikin math

last one today since I just got it and want to write it before I sleep and forget. I copy the test and put comments on italics, ok?

Figure c) demonstrate how to layout the lateral pitch on the top surface of the hip rafter. The plane BCD on figure c) is parallel to O2B1G in figure b), that is, the roof plane which is at 45 degrees is parallel to the chamfer made on the rafter. That much is clear since you are going to nail boards on it and last time I saw boards are flat and go parallel to themselves. Oh, by the way, why not to draw the rafter in a complete different direction in figure b) and c)? Yeah, that's a great idea, otherwise is too simple. So remember, the rafter in c) goes down to the right, and in b) it goes down to the left. Great. The lateral pitch of the top surface is given by the ratio of EF/BF (the rise and run seen from the perpendicular cut of the rafter) on figure c). AD is the centre line of the hip rafter btw.

Assume AC = h. Since C is an arbitrary point we can take any triangle whatsoever, this is a good one. 

CB = 2h. What? Oh yeah, the slope there is 5/10 or 1/2. The line AB goes at 45 degrees.

AB = Sqrt(5)h. Five? Oh yeah, Sqrt(2^2 + 1^2) = Sqrt(5)

CD = 2Sqrt(2)h. That's easy no? The line CD is 45 degrees from DB

AD = 3h = Sqrt( 2^2 sqrt(2)^2 h^2+h^2) since to go from A to D you can go via AC and CD

BD = 2h Seems that CBD form a 90 degrees angle doesn't it?

AB^2 -BE^2 = AE^2 So this is 90 degrees on the chamfer plane

And the rest follows clearly...

if not so clearly, you apply pitagora's theorem to each and every triangle formed. We are interested in the BFE triangle but for that you need first ED, then EF. F' is used so you can use pitagoras theorem on each part of the triangle and to show that BFE is actually a right angle and then you can get the slope with happens to be 1/3. 

It's not difficult but confusing, at least for me the last time I called a triangle for its points as in ABC was 15 years ago in high school.   

So, do you need to know this? Not really. You could make 153 roofs with a regular slope and simply jigging the chamfer at 1/3 all the time. But what if you change the angle? The steps are the same, just replace the sqrt(5) and all the other numbers by what corresponds there.

I know this is an incomplete description, but part of the process is doing the thinking yourself, watching a video of how to use an axe is not learning to use an axe. Same with geometry.

Off to bed, got a cold with this lovely weather and feel like shit. Tomorrow will read the jack rafter section and continue packing my bags, we are off this land in 7 days.

Sashigane addiction

Travelling again, enjoying the German autobahn and it's punctual train and lovely buses. Irony of course, god I'm starting to hate this country.

We visited first Pauli in Essen, then moved to Münster to plane some oak and "Platane" with Till (Pauli's bf) and then I was picked up to help a friend with some restoration work in his family farm. He said it had to do with wood so I said yes, but he didn't mention powertools and dust, then I said no and spend saturday in the kitchen making pizza and sunday mostly in bed playing with my sashigane. Not in a masturbatory way that is.

I didn't bring my computer and my fingers were itching for writing some posts about this magical little tool and all the stuff you can make with it.

It seems to me that this sashigane thing will go for a long time, together with the mathematics/geometry of roof explained by a phisicist so I'm making a new tag for it: sashigane.

Let's go to it then.

First picture: How to draw a given slope. Say 5/10. You take the edge of timber, mark 5 on one side, 10 on the other, and you got it.

If you don't get it the why (as I didn't) maybe you can use some trigonometry.

The two triangles, the one with sides 5 and 10m and the other with sides 10cos(alpha) and 10sin(alpha) are equivalent, they both have the same slope. Homework, prove this.

Now, for Gabe's last joinery layout the half-roof pitch was not explained so here it goes.

Since the rafter is angled at 45 degrees the run becomes 10sqrt(2) You take a perpendicular to it that connects with the edge (same length since it's at 45 degrees) and the hypotenuse of that triangle is the new run for the side cut of the keta. Look here if you don't know what I'm talkin about:

http://granitemountainwoodcraft.com/2015/10/17/layout-for-simple-japanese-hip-roof/

and check this picture in particular:

In Gabe's case the slope is not touching the edge but it's parallel to the line I draw, thus has the same slope. Questions just ask, I don't know your level of geometry so I assume you got the basics as pytagoras, thales theorems, and sin, cos, tan notions.

So, why addiction? well look at this. You just move your sashigane around and the joints start to appear by themselves. Crazy shit.

 Here I was playing with slopes and realized that the splayed leg problems is trivial with a square and some approximations.

So this is the second lesson of today, and it's how to compute the square root of a number without a computer (great, written in a computer, I see the irony).

So, first of all why you need to compute square roots? Because we live in an eculidian space in the local universe. What do I mean? That the shortest line between to points on the world is equal to the square root of the sum of the displacement in each dimension squared. (That is, Pytagoras theorem is not a theorem but an axiom about what kind of world we live in. In you draw a triangle on the surface of a sphere instead of a plane it does not hold anymore. {I told you that learning geometry from a physicist was crazy, they always take detours to explain things.})

Anyway. Sqrt(100) = 10. That is, 10*10 = 100. Sqrt(156) = ????

Let's say we have a splayed leg with the following diagonal slope of 8/10.

First we need to know the length of the leg eh? So by pitagoras the hypotenuse is sqrt(10^2+8^2).

EDIT: thanks to jason I found my math was horrible. 8*8 is not 56 but 64, so now I edit the approximation accordingly. sorry for that.

Exactly, sqrt(164). 13 times 13 is 169. So let's write it like this: Sqrt(164) = Sqrt(169*(1 - 5/169)) = Sqrt(13^2)*Sqrt(1 - 5/169) = 13 * (1 - 0.015)

CHAN!

The last equal sign is not actually an equal sign but an approx. I use the following formula: sqrt(1+x) = 1 + x/2 when x is way smaller than 1. That's called Taylor expansion but doesn't matter a ball the name, just bear with me or google it.

So now that we know the length of the hypotenuse we can compute how much to decrease the square leg so once splayed appears with a square cross section on the plan.

Say we have a 10x10 cross section beam. The new diagonal length needs to be 10/13 so when it's splayed it's again 10.  Let's write 10 = 13 - 3 so 10/13 = 1 - 3/13  = 1 - 1.5/13 - 1.5/13

So you divide the side in 13 parts by using your sashigane in diagonal so it covers 130mm and mark one vertical line every 10mm then you take 1.5cm from each side and pass it to the diagonal with compass and not a vertical line as I did in the drawing.

Now go and try it for a slope 7/10 and tell me the answer. I go and try to cut a gothic roof in the meantime.

Thanks Jason for catching the error and hope this solve your doubts. I may draw the triangles again so they show the right numbers