**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

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