Stationing question

[CPStudent]

I need to find Point A on the following horizontal curve.

I can find PC, PT, M, E, etc. But I can't for the life of me figure out how to find the stationing of Point A.

I calculated the arc length to be 1151.92 feet (to save time in trying to help me).

Any guidance is appreciated. Thanks!

 

[sac_engineer]

Ngnrd is exactly right. The length of the curve is 1162.4 (not sure how cpstudent got 1151.92 feet).

Using 150 feet as the middle ordinate distance (M) for the arc cut by the railroad, the length is 1212.9 feet.

I = deflection angle

R = radius = 1200 feet

M = R ( 1 - cos (I/2) ) ... solve for I given M = 150 feet and R = 1200 feet... I = 57.91 degrees

L = RI/57.29578 = 1212.9 feet

Divide L by two (i.e the arc length between the RR Xing and the PT), you get 606.4 feet. Subtract that from the length to get the arc length from PC to the RR Xing: 1162.4-606.4 = 556 feet.

 

[CPStudent]

Thank you both. The way I was approaching it was that I needed to find the arc length from Point A to PT and subtract that value from the stationing at PT, but I didn't see the part of the diagram where it specified the distance from the railroad is 150 feet.

I instead made a triangle from the origin of the radius to PT to Point A. Because I know O-PT is 1050 feet (1200 - 150), and O-A is 1200 feet, I could calculate the angle between points A and PT, and from there could calculate the arc length from Point A to PT.

 

[CPStudent]

Ngnrd is exactly right. The length of the curve is 1162.4 (not sure how cpstudent got 1151.92 feet).
Using 150 feet as the middle ordinate distance (M) for the arc cut by the railroad, the length is 1212.9 feet.

I = deflection angle

R = radius = 1200 feet

M = R ( 1 - cos (I/2) ) ... solve for I given M = 150 feet and R = 1200 feet... I = 57.91 degrees

L = RI/57.29578 = 1212.9 feet

Divide L by two (i.e the arc length between the RR Xing and the PT), you get 606.4 feet. Subtract that from the length to get the arc length from PC to the RR Xing: 1162.4-606.4 = 556 feet.

sac_engineer,

I calculated the arc length as the angle of the curve x the radius of the curve.

The angle of the curve is 55 degrees. In radians, that is .959931.

So 1200 x .959931 = 1151.92 feet.

How did you get 1162.4 feet?

Thanks.

 

[RIP - VTEnviro]

I think I can help.

 

[civilized_naah]

I think I can help.

I guess he was just asking for it.

 

[civilized_naah]

I need to find Point A on the following horizontal curve.

See attached image.

 

[ptatohed]

I'm confused. You guys are getting 555.85 feet from PC to Point A but, per the graphic, Point A is clearly upstation of the midpoint on the curve. The midpoint is 581.20 feet from the PC. Shouldn't the distance from PC to Point A be greater than 581'? Or is it just a graphical thing?

 

[canvasbak66]

I solved it with two different curves. One with a radius of 1200 ft the other with a radius of 1050 ft. (1050 = 1200 - 150) The difference in L between the two curves is added to L/2 of the original problem (with R= 1200). Compute T and find STA at PC. With that I computed a STA @ Point A = 183+22.67.

 

[civilized_naah]

I'm confused. You guys are getting 555.85 feet from PC to Point A but, per the graphic, Point A is clearly upstation of the midpoint on the curve. The midpoint is 581.20 feet from the PC. Shouldn't the distance from PC to Point A be greater than 581'? Or is it just a graphical thing?

The geometry is not changed by the fact that the point A is before the midpoint (a fact that we do not know at first). The graphic will look different but the results should remain the same.

 

[ptatohed]

I'm confused. You guys are getting 555.85 feet from PC to Point A but, per the graphic, Point A is clearly upstation of the midpoint on the curve. The midpoint is 581.20 feet from the PC. Shouldn't the distance from PC to Point A be greater than 581'? Or is it just a graphical thing?

The geometry is not changed by the fact that the point A is before the midpoint (a fact that we do not know at first). The graphic will look different but the results should remain the same.

Then I wonder what the author's point of dimensioning and labeling 'M' and 'E' was? It served no purpose. I thought the author was trying to clearly demonstate that point A was after the curve's midpoint.

 

[canvasbak66]

I'm confused. You guys are getting 555.85 feet from PC to Point A but, per the graphic, Point A is clearly upstation of the midpoint on the curve. The midpoint is 581.20 feet from the PC. Shouldn't the distance from PC to Point A be greater than 581'? Or is it just a graphical thing?

The geometry is not changed by the fact that the point A is before the midpoint (a fact that we do not know at first). The graphic will look different but the results should remain the same.

Then I wonder what the author's point of dimensioning and labeling 'M' and 'E' was? It served no purpose. I thought the author was trying to clearly demonstate that point A was after the curve's midpoint.

Point A is not before the midpoint.

 

[civilized_naah]

Alright, here's a scale drawing using the dimensions given in the original diagram. As you can see, Point A is clearly back-station of the PC - PT mid point.

We had 2 answers for location of point A - differing by 0.11 ft. In one solution the arc length to A was 555.96, in another 555.85. I could not discover where the difference came from. What is the exact location from your CAD drawing?

 

[civilized_naah]

We had 2 answers for location of point A - differing by 0.11 ft. In one solution the arc length to A was 555.96, in another 555.85. I could not discover where the difference came from. What is the exact location from your CAD drawing?

Station of Point A from my Cad drawing is 181+52.1262. Arc length from PC to Point A is 555.9567.

The difference is due to the fact that your method measures a straight line distance from PC to A (law of sines only applies to triangles, which have straight sides), while my method measures the arc length between them (a curve). And, of course, the shortest distance between two points is a straight line, which is why your measurement is slightly shorter than mine.

Actually, I figured it out. It was just the precision to which I calculated the deflection angle alpha. In subsequent steps, I input 13.27 degrees, whereas a more precise value is 13.2725 degrees. If I use the latter the arc length works out to be 555.9574. There is no approximation (arc vs chord) in my solution, so that's OK.

 

[ptatohed]

Gosh. I like to think I am good at these geometric design problems but this one stumped me. :huh: