It's been a while since I've seen one of these discussed. I think I have this one worked, and would like to compare answers with anyone else. It's late enough in the month that I don't suppose anybody is going to use our discussion to win the contest.
The only assumption I made is that the curve is tangent. I thought for quite a while to come up with an approach that simplified things a little, and there is still a fair amount of calculation.
Okay....I'll try.
Assuming Pt. 1 is the PC:
Angle x =13-29-28
c-1 Length=117.773
I drew it up in Carlson and came up with 13-29-43 & 117.77'
We agree very closely. I had 13-29-43 and 117.770 using Excel to do the calculations as I worked through coordinates for lots of points.
I found it easier to ignore the given directions and just keep the 10 deg angle, and compute with a coordinate system where their N70E line was parallel to my X axis, and drew in a couple right triangles.
Did you find an easier method of attack?
Me and the old HP 48 came up with 13°29'43" for angle "X" and c1 = 117.7684
I ended up using the law of sines quite a bit, but something similar to what you were saying Bill. I solved for 4 coordinates. The coordinates were the check on the triangles for me.
I drew it up in Autocad 2004 and came up with 13-29-43 & 117.77'
I did it without coordinates.
Draw a line from pt. 2 perpendicular to the lines emanating from pt.5 & pt.3, crossing the line from pt.5 at pt.8 and crossing the line from pt.3 at pt.7
Angle 1-2-8 is 20°, angle 1-2-3 is 22°55'06"
Distance 2-7 is 500xcos32°55'06"=419.723, making distance 2-8 = 344.723 and therefore angle 5-2-8 = arccos(344.723/500) = 46°24'49" and angle 5-2-3 = 13°39'43" and c-1 = 117.7695
> We agree very closely. I had 13-29-43 and 117.770 using Excel to do the calculations as I worked through coordinates for lots of points.
>
> I found it easier to ignore the given directions and just keep the 10 deg angle, and compute with a coordinate system where their N70E line was parallel to my X axis, and drew in a couple right triangles.
>
> Did you find an easier method of attack?
Here's how I and my nine-dollar Sharp EL-501W worked the problem:
AutoCAD all the way for me. It's where I do pretty much all surveying calculations except for GPS baseline processing and least-squares adjustments.
> AutoCAD all the way for me. It's where I do pretty much all surveying calculations except for GPS baseline processing and least-squares adjustments.
Yes, but how do you fit it into your shirt pocket? :>
Kent and Dave have much better approaches than I used. I constructed one of the triangle Dave did, but got focused on coordinates and didn't think of using both triangles like he did.
Thanks, guys.
20 degrees for angle 1-2-8 is a typo, maybe? The sum in the next line seems to make it 10, which makes sense in other ways, it seems to me. But you guys are all way ahead of me.
Cheers,
Henry
Oops!
You're right. I mistyped it. Angle 1-2-8 is 10°.
Bill, i solved for I of 200' curve (22.5506), calcd bearing of middle radial line, det 'top' angle of bottom-most triangle(2,5,2-3); drew right triangle at western end of 75' parcel, calcd the two distances on the middle radial line (89.345 & 410.655), used sine law to det angle opp angle X (@ 5) and thus got angle X and dist c-1 - which matches everyone else's
Good job guys....Now try it changing the beginning coordinates to 10,000/3,000 😉
Don't think it would matter since you didn't travel West more than a couple hundred feet.
Changing it to 0,0 would likely cause more problems, if you wanted to go the coordinate route.
Kris....maybe you missed my smiley-face. You could change the coordinates to anything; and it wouldn't affect the answer.
Okay....bad joke. If you have to explain the joke, and/or if no one laughs, it probably wasn't funny.:-P
it got a chuckle out of me 😀
