*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.

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