Makings of Edges
DRAWING
Which gives us then 2 into 75 now unto men
100
25
______
75
2 into 75
3
__________
6
__________
15
7
___________
14
___________
10
5
____________
10
____________
0
Which would be 37.5 on each side with 25 contained in the middle. Now how is that possibly in solving for figure? The other side then since it is shorter can bear with the 20 some on the side of the 50.
DRAWING
50
20
_____
30
2 into 30
1
_________
10
5
__________
10
__________
0
There, that is 15 on each side of the 20. If you have decided safety was not as important quite as looks in the matter, we would run it again just to see what would happen with the slightest of beams. I can either run them this way or that way from there once again and probably both ways just to enusre there is no collapse at the end. If trees though came expressed as the slightest of these, the longest possible numbers would be 50 or 100 foot timbers for the roof then in sum.
50
25
___
25
2 into 25
1
________
5
________
2
________
4
________
10
_________
5
There, that is 12.5 foot on the edges or cubits, rather whatever. If you had remembered before it matters a lot getting your measures mixed up for sure.
100
20
___
80
2 into 80
4
________
80
_______
0
40 times at the end. Alright, now let us stack it all like this in the figure.
DRAWING
By dividing the 50 and 100s in half in the middle, we can figure out the least common determinant safety in factor.
2 into 50 goes 25 times
2 into 100 goes 50 times
Alright so if you suspended a beam in the air with nothing underneath it general then walked atop ithad most likely snap in the middle. If I had just one support right in the middle, it is gonna break when you get towards the sections. If you put two supports too close together, it would break on the edges in general. You can already tell that the 40s wo not be working at all on the edges. Ihad estimate that about 10 or so on each side would end up in breaking. It only follows naturally that you would be looking at
DRAWING
As the best possible factor.