How to calculate station and offset on a curve

On a curve, would the offset align with the radius point and the point of the original alignment?
>

Yes

It would also probably equal the deflection angle plus or minus 90°

> It would also probably equal the deflection angle plus or minus 90°

You know better than that.

Think about it. B-)

> > It would also probably equal the deflection angle plus or minus 90°
>
> You know better than that.
>
> Think about it. B-)

I.B Thinkin , what am I missing?

> > > It would also probably equal the deflection angle plus or minus 90°
> >
> > You know better than that.
> >
> > Think about it. B-)
>
> I.B Thinkin , what am I missing?

From the PC to a station is the Deflection angle for that station, plus at that station the deflection angle again ± 90°.

Upon reflection, it may be in the way we each picture the problem.B-)

Yes sir. I was thinking if you had a deflection of 12°..

0°+12°+90°=102° to get to the outside. If you flopped, turned, whatever the 180° to get to the inside you have °282. 90°-12°-360°= 282°

Sometimes I miss the good old days:-(

I wish I had a company that would pay for software that could calculate this for me.
Well its good to know how to do it too I guess.

So how do you find the deflection angle? I do have all the curve data (PC PT coordinates and station, delta, direction, L, E)

Length of curve divided by the radius times 200 equals the deflection in degrees, minutes and seconds to any point on curve if you are at PC back-sighting rear tangent and adding 180 degrees for curve to the right. If to the left, subtract from 180 degrees.

If you have coordinates, inverse PC to RP and RP to the point. The distance RP to point minus the radius is your offset. The angle PC-RP-Point is the delta for figuring the station.

This isn't exactly what you're looking for, but might be useful. It calculates the curve formulas from any reasonable combination of parameters.
https://dl.dropboxusercontent.com/u/25124076/CurveCalc.exe

There's no warranty, so check it with known examples before relying on it.

> So how do you find the deflection angle? I do have all the curve data (PC PT coordinates and station, delta, direction, L, E)

Degree of Curve divided by 0.3 = factor to use to obtain deflection for any length of arc on center line.
i.e. 1° Curve/.3 = .30 factor. ---- 50’ L x.30 = 15’ deflection angle --- 100’ L x .30 = 30’ deflection angle

2° Curve/.3 = .60 factor ---- 50’ L x .60 = 30’ deflection angle ---- 100’ L x .60 = 1° deflection angle

Chords on both the inner and outer offsets to match the center line stations can be calculated using:
2RxSin½ Delta (the deflection angle), and if feasible, then the offsets can be staked using the same deflection angles as used staking the center line from the P.C.

Been a long time, but if I remember correctly, occupying each center line station and setting the station deflection angle on the plates (the correct side of zero of course) and back sighting the P.C., then adding 90° to the occupied station deflection will be to the outer offset line. Just flop the scope to get the inner offset line point. This should be radial to the curve.