TI-84 Plus CE and Casio fx-5800P: Highway Transition Spiral
Introduction
For any given parameters of a highway transition spiral:
PI: point of intersection of the vertices (X coordinate, Y coordinate; Program Variables A, B)
∆: intersection angle (Program variable I)
D_C: degree of the curve (Program variable D)
L_S: length of the spiral (Program variable L)
Outputs:
Z: Angle between the radii of spiral at TS and SC
R: radius of circular curve
TS: point of intersection of main tangent and approach spiral (Program Variables U, V)
SC: point of intersection of main tangent and circular curve
ST: short tangent
Formulas:
Z = L * D / 200
Z is in degrees
Convert Z to radians: Q = Z * π /180
x = L * (1 - Q^2/10)
y = L * (Q/3 - Q^3/43)
k = x - R * sin Z
p = y - R * (1 - cos Z)
T = (R + P) tan (∆/2) + K
S = Y * csc Z = Y / sin Z
Coordinates of TS:
[ PI_X - intg(T/100), PI_Y - 100 * frac(T/100)]
Coordinates of SC:
[ TS_X + intg(L/100), TS_Y + 100 * frac(L/100)]
TI-84 Plus CE Program HYSPIRAL
(program to be typed)
Degree
"EWS 2019-08-03"
Disp "HWY SPIRAL"
Input "PI X: ",A
Input "+ PI Y: ",B
Input "INT-ANGLE: ",I
Input "LENGTH: ",L
Input "DEGREE: ",D
L*D/200→Z
100/(D*π/180)→R
Zπ/180→Q
L(1-Q²/10)→X
L(Q/3-Q³/43)→Y
X-R sin(Z)→K
Y-R (1-cos(Z))→P
(R+P) tan(I/2)+K→T
Y/sin(Z)→S
Disp "ANGLE TS-SC:",Z,"RADIUS:",R
Pause
A-iPart(0.01T)→U
B-100 fPart(0.01T)→V
Disp "TS:",U,"+",V
Pause
Disp "SC:",U+iPart(0.01L),"+",V+100 fPart(0.01L)
Pause
Disp "ST:",S
Casio fx-5800P Program HWYSPIRAL
Deg
"HIGHWAY SPIRAL"
"PI X: "?→A
"PI Y: "?→B
"INT-ANGLE: "?→I
"LENGTH: "?→L
"DEGREE: "?→D
L*D÷200→Z
100÷(D*π÷180)→R
Z*π÷180→Q
L*(1-Q²÷10)→X
L*(Q÷3-Q^(3)÷43)→Y
X-R sin(Z)→K
Y-R (1-cos(Z))→P
(R+P) tan(I÷2)+K→T
Y÷sin(Z)→S
"ANGLE TS-SC:"
Z⊿
"RADIUS:"
R⊿
A-Int(0.01T)→U
B-100 Frac(0.01T)→V
"TS X:"
U⊿
"TS Y:"
V⊿
"SC X:"
U+Int(0.01L)⊿
"SC Y:"
V+100 Frac(0.01L)⊿
"ST:"
S
Example
A highway with a transition spiral is at station 50 + 64.84, with the intersection angle 50°, and the degree of the curve at 6°. The length of the curve is said to be 360 ft.
Inputs:
PI X = 50
PI Y = 64.84
INT-ANGLE = 50
LENGTH = 360
DEGREE = 6
Outputs:
ANGLE TS-SC = 10.8
RADIUS = 954.9296586
TS: (X + Y) = 44 + 37.12990632
SC: (X + Y) = 47 + 97.12990632
ST = 120.4143341
Source:
Hicks, Tyler P.E. Handbook of Civil Engineering Calculations McGraw Hill: New York. 2000 ISBN 0-07-028814-3
Eddie
All original content copyright, © 2011-2019. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.
Introduction
For any given parameters of a highway transition spiral:
PI: point of intersection of the vertices (X coordinate, Y coordinate; Program Variables A, B)
∆: intersection angle (Program variable I)
D_C: degree of the curve (Program variable D)
L_S: length of the spiral (Program variable L)
Outputs:
Z: Angle between the radii of spiral at TS and SC
R: radius of circular curve
TS: point of intersection of main tangent and approach spiral (Program Variables U, V)
SC: point of intersection of main tangent and circular curve
ST: short tangent
Formulas:
Z = L * D / 200
Z is in degrees
Convert Z to radians: Q = Z * π /180
x = L * (1 - Q^2/10)
y = L * (Q/3 - Q^3/43)
k = x - R * sin Z
p = y - R * (1 - cos Z)
T = (R + P) tan (∆/2) + K
S = Y * csc Z = Y / sin Z
Coordinates of TS:
[ PI_X - intg(T/100), PI_Y - 100 * frac(T/100)]
Coordinates of SC:
[ TS_X + intg(L/100), TS_Y + 100 * frac(L/100)]
TI-84 Plus CE Program HYSPIRAL
(program to be typed)
Degree
"EWS 2019-08-03"
Disp "HWY SPIRAL"
Input "PI X: ",A
Input "+ PI Y: ",B
Input "INT-ANGLE: ",I
Input "LENGTH: ",L
Input "DEGREE: ",D
L*D/200→Z
100/(D*π/180)→R
Zπ/180→Q
L(1-Q²/10)→X
L(Q/3-Q³/43)→Y
X-R sin(Z)→K
Y-R (1-cos(Z))→P
(R+P) tan(I/2)+K→T
Y/sin(Z)→S
Disp "ANGLE TS-SC:",Z,"RADIUS:",R
Pause
A-iPart(0.01T)→U
B-100 fPart(0.01T)→V
Disp "TS:",U,"+",V
Pause
Disp "SC:",U+iPart(0.01L),"+",V+100 fPart(0.01L)
Pause
Disp "ST:",S
Casio fx-5800P Program HWYSPIRAL
Deg
"HIGHWAY SPIRAL"
"PI X: "?→A
"PI Y: "?→B
"INT-ANGLE: "?→I
"LENGTH: "?→L
"DEGREE: "?→D
L*D÷200→Z
100÷(D*π÷180)→R
Z*π÷180→Q
L*(1-Q²÷10)→X
L*(Q÷3-Q^(3)÷43)→Y
X-R sin(Z)→K
Y-R (1-cos(Z))→P
(R+P) tan(I÷2)+K→T
Y÷sin(Z)→S
"ANGLE TS-SC:"
Z⊿
"RADIUS:"
R⊿
A-Int(0.01T)→U
B-100 Frac(0.01T)→V
"TS X:"
U⊿
"TS Y:"
V⊿
"SC X:"
U+Int(0.01L)⊿
"SC Y:"
V+100 Frac(0.01L)⊿
"ST:"
S
Example
A highway with a transition spiral is at station 50 + 64.84, with the intersection angle 50°, and the degree of the curve at 6°. The length of the curve is said to be 360 ft.
Inputs:
PI X = 50
PI Y = 64.84
INT-ANGLE = 50
LENGTH = 360
DEGREE = 6
Outputs:
ANGLE TS-SC = 10.8
RADIUS = 954.9296586
TS: (X + Y) = 44 + 37.12990632
SC: (X + Y) = 47 + 97.12990632
ST = 120.4143341
Source:
Hicks, Tyler P.E. Handbook of Civil Engineering Calculations McGraw Hill: New York. 2000 ISBN 0-07-028814-3
Eddie
All original content copyright, © 2011-2019. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.