First thing this morning, while everyone stood around waiting (and before I had coffee), I had to solve this little puzzle. All I had was my construction master trig calculator.
The big deal was installing retail storefronts between existing columns.
I admit, I got it wrong the first time I calced it, but caught my mistake on the second try.
The black lines are column lines.
The blue line is a reference line we used to align the arches several years ago.
The "grey" arc is 71.7 degrees.
The red symbol is an "arch" column, seven inches wide on each side of the "N-S" column line.
The yellow dimension is 14".
The green dimension is 2'-8".
The question I needed to answer was, how much and in which direction does the blue reference line need to move in order to be 4 feet north of the "E-W" column line.
Forgive the sketch, I used the first online sketch pad I found.
0.22' north?
The blue reference line needs to be moved 1/2" north or 0.0424' north.
I’d sure like to see the solution to that. I have no idea how Roadhand and Floyd came to their conclusions, but the problem whizzed way over my head! But hey! I’m an old fart now, so I can use that as an excuse as to my ignorance. 😉
Y'all have a great weekend! 🙂
> The big deal was installing retail storefronts between existing columns.
Does it really matter if you want to use your old data? You have to fit what's there.
Being existing conditions, I'd tend to just take my handy Lufkin out and measure 4 ft from the outside most columns (or arches, or whatever they are) to establish the line. Then check a few in the middle and hope they are within tolerance.
But ballpark it seems it would be around 0.8'.
The blue line needs to move 2 11/16 " perpendicular to the E-W axis line.
("North" on the sketch.)
Law of sines dictates about 0.23'
I thought this was a bit of fun...maybe this will help.
I rushed through the sketch and it showed, so in all fairness, I thought I should clarify it a little.
I'll post my solution and method tomorrow morning.
I thought this was a bit of fun...maybe this will help.
Is the end of the yellow dimension to the same "line" or point that the green dimension comes from?
0.0425 North.
I thought this was a bit of fun...Answer
OK, sneaky little bugger.
First, I took the 2' - 8" and converted it to 2.667'. I need that dimension as it is.
Next I look at the yellow. The 14" is from the cl x cl (of the column lines)to the NE corner of the column. So, I need to solve that for the hypotonuese.
So 1' - 2" equals 1.167 and that's the base.
And all the triangles are right triangles with 71.7 deg and 18.3 deg as the angles.
So, I need h. Well, h = b/sin B or, h equals 1.167'/sin 71.7 deg.
Or, h equals 1.229'.
I need that dimension.
Now I need to know if I solved to the full dimension of the half width of the column, or 7".
Seven inches equals 0.583'
So if a^2 + b^2 = c^2, then a^2 = c^2 - b^2.
Thus a = 0.385'
And 0.583' - 0.385' = 0.198'.
I have another triangle to solve whose a = 0.198'.
I need the height of this triangle, and I have one side and two angles.
So b = a sin B / sin A.
Well A is 90 deg and the sin of 90 deg is 1, so I toss it out.
Thus b = a sin B, or b = 0.198' sin 18.3 deg.
B = 18.3 deg, so b = 0.062'
Now I have all the dimensions I need.
2.667' from the green.
1.229' from the black triangle.
0.062' from the blue triangle. And that equals;
3.959'
I need 4.000', so I need to move the blue reference line 0.041' feet north to be 4' - 0" from the east/west column center.
I thought this was a bit of fun...Answer
Ah-ha......I missed the fact that the yellow dimension was from the intersection of the axis lines to the corner of the column, not the length along the easterly face of the column.
