How to draw reverse curves between parallel tangents

You need to know where your begin of curve is on one of the two tangents.

1. Offset a line 500' towards the radius point from each tangent.

2. The point on the 500 foot line per 1 above opposite the BC point is the radius point of a 1000 foot radius circle.

3. Where the 1000 foot radius circle intersects the other 500 foot offset line is the radius point opposite the other BC.

Thanks man, my brain just hit a brick wall today.?ÿ Here is TS response too (very similar):

I don't know of an automated way to do this - I'd offset one of the lines by 25' and intersect the line between the center of the two circles - then draw a tangent curve from each line to the intersection of the above line (for the tangent curve, just trace over the existing polylines and enter the curve mode at the pc's).

Here is how I do it. Changed the radius on one curve to simplify trying to describe it.

?ÿ

After looking at Karoly's drawing below, It seems I misunderstood the question ?????ÿ

Had an old time engineer show me a shortcut method to solving this math problem. ?ÿWish I would have pegged all those old shortcuts ?ÿ I once used to pass on to curious minds for old time sakes. ?ÿMaybe someone will post it. ?ÿGood day, Jp

Basic math:

Set up a 500 ft circle as DK shows. Draw a horizontal center line between the given parallels.?ÿ This cuts off the north-heading radius at 475 ft.?ÿ Draw another radius to the intersection of the circle and the center line.

Now you have a right triangle 475 side, 500 hypotenuse.?ÿ Solve for other side 156.125 (centerline distance to PCC) and delta angle 18^ 11' 41.5"?ÿ That defines the curves.?ÿ I get L=158.780

@bill93

I dont know if Hypotenuse, Delta angle, radius, would be called basic math,?ÿ think it is Pure mathematics, hopefully Mathteacher can let us know.

@bill93

I'm not sure what you mean by the 475' side but

I get:

delta = 18deg11'42"

Arc L = 158.78'

T = 80.06'

Ch = 158.12'

If the tangent in is East then the chord bearing is S 80deg 54'09" E.

My method: You have four parallel lines, the original two in the middle 50' apart and one 450' north of the north middle line and one 450' south of the south middle line. Adding 450' (south to south middle) plus 500' (south middle to north) is 950'. My triangle is south r.pt. 1000' to north r.pt. (hypotenuse) then 950' south to south line then this yields 312.25' back to south r.pt. by Pythagorean Theorem. Arc-sin 312.25/1000 equals 18deg 11'42" delta. After that it's just curve formulas.

My sketch to go with above description

@bill93

I like it, never thought of that way.

My initial post above was more a description of how to do it in CAD.

Here's my updated sketch: