Determining my Chord bearing and distance

Chord length is easy: arc length/radius = delta (in radians) x180/pi= delta (in degrees)

chord is 2*radius*sin(1/2delta) = 116.31'

I don't see enough information to compute the bearing. There's not a tangent or radial bearing expressly shown there, and there isn't enough info on the remaining sides of the property to cogo around the lot to deduce it. Possibly if we had the entire document there might be a tangent bearing further to the west on the railroad right of way that might be the key to get there, but I'm not betting on it. It appears that there is a good chance that the road right of way and the railroad right of way intersect while both are still in a curve, so unless there's a tangent, radial or chord bearing somewhere off the page I don't see a lot of options to get there. Another trick is to look in the legal description (if there's on on the face of the document), it may have info not shown in the graphic representation.

Carlson and I would assume most other software can build curves off the info given.?ÿ You'd have to start somewhere on the survey where a curve comes off a straight line and make the assumption that the curve is tangent to the line you're starting with.?ÿ Carlson even has a routine that can calculate reverse curves where you transition from one curve to another with no lines in between.

?ÿ

@david-baalman

The radius of the arc highlighted by the OP is shorter than the radius of the cul de sac, so I think you're not on the right track.

This smells like a student fishing for a test answer.?ÿ I'll tinker with it later, maybe.?ÿ ?????ÿ

With only the information shown, the chord bearing cannot be calculated. Seeing the entire map could possible reveal missing info, as others have said.

@bstrand I don't see what's leading you to that conclusion. I see a leader with curve info (including a 60' radius) pointing at the curve in question, similar to the other curves we can see on the screen grab.

@david-baalman I believe you could calc the chord length but not the bearing with the info shown.?ÿ I don't know of a way to calculate the chord bearing without a tangent line to work off of.

@david-baalman

Right, but what makes you think the pins that are anchoring the arc in question are on exact opposite sides of the circle??ÿ In fact, some quick math will show they aren't.

2*pi*r = circumference, so for this circle 376.99.?ÿ Half of that dimension would be across from each other which is 188.49.?ÿ The dimension between the pins according to the plat is 158.68.

@bstrand I didnƒ??t say that they were. If youƒ??ll look at the chord distance comp I showed, youƒ??ll see I computed the delta angle based on the arc length given. I compute that delta as 154-31-42 (dms) which yields 116.31ƒ?? for a chord distance as I stated. A 180 delta would be a 120ƒ?? ,chord obviously.?ÿ

@david-baalman

OK, but you used a radius of 60 feet to compute your delta and that only works if the pins in question are on exact opposite sides of the circle which we know aren't.?ÿ Since we know the distance along the arc is less than the 188.49 required for a 60 foot radius then the radius has to be less than 60 feet which makes your calculation incorrect.

Think about it this way, if you drew a straight line (chord) between the pins right now and drew a line from the middle of that chord to anywhere on the arc it's impossible to get 60 feet.

@bstrand youƒ??re saying that there is no 60ƒ?? radius curve that can result in an arc length of 158.68 feet?

If you can compute the radius point for the center of your arc, the chord bearing will be the bearing between the radius point and the PC, plus 90 degrees to get the incoming tangent bearing plus half the delta of the curve to get the chord bearing.?ÿ

@david-baalman

No, I'm not saying that all; a 60 foot radius curve with a length of 158.68 feet would have a delta of 151?ø31' 41" and not 183?ø33'30" as is shown on this plat.

The key to all of this is the arc formula in your very first post.

As far as the bearing, yeah it looks like the arc is non-tangent to all of the lines who's bearing is known so the chord bearing is probably a lost cause.

EDIT:?ÿ I say probably because I had made that assumption once or twice in school and been burned when a bearing from across the map is rotated, translated, offset and blah blah to get the bearing of the line I was working on. ??? So it wouldn't surprise me if something like that was applicable here... because like I say this looks like a test or homework problem if I've ever seen one.

?ÿ

@bstrand the delta of 183-33-30 is the overall delta of the curves across lots 3 and 4 up to the reverse curve on lot 4

If the OP could provide the bearing and distance of the short, straight boundary on the bottom of Lot 3, it would be possible to work around Lot 3 to nail down another point on the cul-de-sac.?ÿ Then the answer should be found.

@david-baalman

Exactly, so how can that also be the measure of the angle between the pins the OP has highlighted?

There are the 3 elements of an arc, if you change one then at least one of the others has to change.?ÿ The plat information describes an arc 192.22 feet long, but we're interested in an arc 158.68 feet long.?ÿ If we want to keep the same radius on the shorter curve then the delta has to shrink, or if we want to keep the same delta then the radius has to shrink.

And this is where the ultimate solution to the problem lies I think.?ÿ I'm pretty rusty on this but I think since pi is involved and is a ratio then it's possible to use proportioning to get the answer.?ÿ I'll fiddle with it after dinner and see.?ÿ If some other math guru has the answer then feel free to throw it out there.

@bstrand simple they gave the delta only on the overall curve and the arc lengths of the individual segments for each arc, but not the deltas for them. Chord distance is the number I gave in the original reply, delta is as computed above as well.

I think we can all agree that the chord bearing is a lost cause without additional info, and on a more soap-boxy note, whoever drew and stamped this plat, short plat, partition plat or whatever it is gave little to no thought about someone having to retrace it in the future, unless there is significant information shown elsewhere on it that we canƒ??t see in this screen grab.?ÿ

@david-baalman

OK, so maybe this will demonstrate the point.?ÿ Take the delta you calculated for the 158.68 foot curve, calculate the delta for the 33.54 foot curve and add them together.?ÿ Do you get 183?ø33'30" like the plat says you should??ÿ I will eat my work boots if you do.?ÿ

DUEL?ÿ At sunrise.?ÿ EDM's.?ÿ No seconds allowed.?ÿ At thirty paces each. The first to drive a nail, set up perfectly over the center of it and level perfectly while drinking a two-liter of root beer wins

@bstrand Within 6 arc seconds yes. 33.54' arc length at 60' radius is a delta of 32d01'42". 158.68' arc length at 60' radius is 151d31'42". Add those up you get 183d33'24", so within 6" of 183d33'30" shown on the plat. With arc length given only to 0.01' that is all just rounding error.?ÿ