Geometric Design Question

[jaa046]

can anyone solve this:

A circular horizontal curve with a radius of 1430' and a deflection angle of 14"30' has its point of intersection at sta. 572+00. The forward tangent is to be shifted parallel to itself and 12' to the left. If the point of curvature station remains in the same place. What is most nearly the radius of shifted curve?

I dont know the answer to this but I got 1810', i just wanted to check if you get the same answer.

Thanks.

 

[civilized_naah]

One missing piece of information is whether the curve itself deflects left or right. If the curve deflects to the right and the forward tangent is offset to its left (inwards), then the radius becomes shorter. If the tangent is shifted to the right, then the radius becomes longer.

The key to this problem is in the triangle formed by the parallel offset (12 ft) and the resulting shortening of the tangent. The shortening of the tangent is delta-T = 12/sin 14.5 = 47.93 ft

Since the deflection angle (I) remains 14.5 degrees, the delta-T = 47.93 ft means a delta R = 47.93 / tan 7.25 = 376.8 ft. Therefore, the new radius = 1430 - 376.8 = 1053.2 ft

If the tangent is shifted outward, the longer radius would be 1430 + 376.8 = 1806.8 ft

 

[Ambrug20]

PC = 570+18.08

T = 193.92 by CERM 78.4

T' = T+12=193.92

R2T' / tg (14.5/2) = 1524.33

see attached file

Hope I am right.

 

[ptatohed]

I get one of c_n's answers (c_n, the problem statement reads "to the left" so I think we can assume that the forward tangent is offset inward if we draw the curve in the typical fashion with the PI pointing "up"). I get R2 = 1053.25'

T1 = R1 Tan (I/2) = 1430' Tan (14.5/2) = 181.82'

New PI sta = 572+00 - 'x'; Sin 14.5 = 12 / 'x'; 'x' = 47.93'

T2 = T1 - 'x' = 181.92' - 47.93' = 133.99'

R2 = T2 / Tan (I/2) = 133.99 / Tan (14.5/2) = 1053.25'

 

[sac_engineer]

My approach is like ptatohed, but my formula's are different:

Chord length of original curve (horizontal line connecting PC and PT) = 2 x 1430 x sin(14.5/2) = 360.92 feet

Since the PT is shifted horizontally to the left by 12 feet and PC is the same point, the new chord length is 360.92 - 12 = 348.92 feet and the new radius ® can be calc'd by ratio with respect to original radius and original chord:

new (chord / radius) = original (chord / radius)

348.92/R = 360.92/1430

R = 1382.45 feet

 

[ptatohed]

My approach is like ptatohed, but my formula's are different:

Chord length of original curve (horizontal line connecting PC and PT) = 2 x 1430 x sin(14.5/2) = 360.92 feet

Since the PT is shifted horizontally to the left by 12 feet and PC is the same point, the new chord length is 360.92 - 12 = 348.92 feet and the new radius ® can be calc'd by ratio with respect to original radius and original chord:

new (chord / radius) = original (chord / radius)

348.92/R = 360.92/1430

R = 1382.45 feet

sac, unfortunately, I don't think this will work because the forward tangent is "offset" 12' from the original forward tangent, not "slid" 12' along the chord. At least that's how I understand these types of problems when the word "shifted" is used.

 

[sac_engineer]

The only info we have is: "The forward tangent is to be shifted parallel to itself and 12' to the left."

The shift can only be assumed a slide 12 feet to the left because no other directional information is provided. If it's a right angle offset, then we would only need to add or subract this offset to the radius.

jaa046: can you provide a diagram?

 

[sac_engineer]

Also, the arc length of the original curve is 362 feet, so a shift of 12 feet of the PT (i.e. 3% of the length), regardless of direction, would not yield nearly a 400-foot difference in the radius (30% difference less than the original) as your answer is showing.

 

[ptatohed]

Also, the arc length of the original curve is 362 feet, so a shift of 12 feet of the PT (i.e. 3% of the length), regardless of direction, would not yield nearly a 400-foot difference in the radius (30% difference less than the original) as your answer is showing.

Please double check your work.

My answer of R2 = 1053.25' is correct if the forward tangent is offset.

And your answer of R2 = 1382.45' is correct if the forward tangent is slid.

Without further information and without a diagram I can only make an assumption that the forward tangent is offset (not slid). I think this is how I have seen other practice problems. Also, a decision to move the tangent is typically to avoid conflict with existing features and those features are typically measured perpendicular to the centerline.

Anyway, hopefully jaa can provide a diagram for this problem.

 

[mrt406]

A circular horizontal curve with a radius of 1430' and a deflection angle of 14"30'

Might want to double check that. Looks like everyone is assuming you meant 14 degrees-30 minutes.... but this is 14 seconds-30 minutes which, apart from being unorthodox, is completely different.

 

[ptatohed]

^^ I noticed that too but just assumed he meant 14 deg 30 min.

I was thinking about this problem on my way home last night and I'm thinking now that maybe "to the left" might mean offset to the left as looking up-station. In this case, civilized's larger radius answer would be the correct one (post #2).

 

[sac_engineer]

I was also thinking about the question and my previous method was incorrect. I agree with ptatohed that the offset to the left is with respect to the direction of the curve and would increase the radius to 1806 feet. The left/right perspective is similar to hydraulic modeling where the left and right banks are located with respect to the flow direction of the stream.

 

[ptatohed]

I was helping my friend at work study for the PE exam and I came across this very problem. It's problem #11 from the Transpo 6 Minute Solutions (2nd Ed). The forward tangent is, in fact, offset to the left, looking up-station, (offset to the outside) increasing the radius. The choices are A 1050' B 1530' C 1810' D 1820'. The answer is C 1810' (1806.8' rounded up). Good job c_n (post #2).

 

[rejectedbytexas]

1806'

 

[ptatohed]

1806'

1806.8' actually. So you bumped a 19 month old thread to provide an answer (with no solution) that has already been provided?

 

[Monkey]

Attached file shows the illustration to the posted question. Some questions are better understood, when there is a sketch to go along with it.

Tangentshift.pdf

 

[ptatohed]

Attached file shows the illustration to the posted question. Some questions are better understood, when there is a sketch to go along with it.

Monkey, dude, you didn't have to re-draw it from scratch. As mentioned earlier, it is from the 6MS books (#11 in the 2nd Ed and #1 in the 3rd Ed).

 

[John QPE]

Is that not a shift to the right? New PI of 572+12?

 

[ptatohed]

Is that not a shift to the right? New PI of 572+12?

No, it is a shift/offset the left. Remember, you should be looking up-station.

And the new PI won't be exactly 572+12 (you are offsetting the forward tangent by 12', not lengthing the back tangent by 12').

 

[Monkey]

Thanks, Ptatohed, I did not have that book. The new PI is (572+00) +( 0+47.93)= (572+48)