Showing posts with label vertical curve. Show all posts
Showing posts with label vertical curve. Show all posts

Sunday, February 17, 2019

HP Prime and TI-86: Minimum Vertical Curve Length


HP Prime and TI-86: Minimum Vertical Curve Length

Introduction

The program MVCL calculates the minimum vertical curve length for sight distances for crest curves (curve that rises then falls) and sag curves (curves that falls than rises). The equations used were determined by the AASHTO (American Association of Highway and Transportation Officials of Washington, D.C.).

HP Prime Program MVCL

EXPORT MVCL()
BEGIN
// Minimum stop speed
LOCAL g1,g2,a,c,l,s,g;

MSGBOX("Break = 2.5 s,
 Decel = 11.2 ft/s^2");

LOCAL l1:={15,20,25,30,35,40,45,
50,55,60,65,70,75,80};

LOCAL l2:={80,115,155,200,250,
305,360,425,495,570,645,730,
820,910};

INPUT({g1,g2,{c,l1}},"MVCL",
{"Grade1%:","Grade2%:",
"Speed:"});

s:=l2(c);
a:=ABS(g1-g2);

l:=2*s-2158/a;
IF s < l
l:=a*s^2/2158;
END;

g:=2*s-(400+3.5*s)/a;
IF s < g
g:=(a*s^2)/(400+3.5*s);
END;

PRINT();
PRINT("Stop speed (ft)");
PRINT("Crest curve: "+l);
PRINT("Sag curve: "+g);


END;

TI-86 Program MVCL
(744 bytes)

Input “GRADE %1:”, G1
Input “GRADE %2:”, G2
abs(G2-G1) → A
Disp “Break time = 2.5s”, “Decl. = 11.2 ft/s²”,
Car Speed?”,”(mph)”

Menu(1,”15”,A,2,”20”,B,
3,”25”,C,4,”30”,D,
5,”35”,E,6,”40”,F,
7,”45”,G,8,”50”,H,
9,”55”,I,10,”60”,J,
11,”65”,K,12,”70”,L,
13,”75”,M,14,”80”,N)

Lbl A : 80 → S : Goto Z
Lbl B : 115 → S : Goto Z
Lbl C : 155 → S : Goto Z
Lbl D : 200 → S : Goto Z
Lbl E : 250 → S : Goto Z
Lbl F : 305 → S : Goto Z
Lbl G : 360 → S : Goto Z
Lbl H : 425 → S : Goto Z
Lbl I : 495 → S : Goto Z
Lbl J : 570 → S : Goto Z
Lbl K : 645 → S : Goto Z
Lbl L : 730 → S : Goto Z
Lbl M : 820 → S : Goto Z
Lbl N : 910 → S : Goto Z

Lbl Z
Disp “Stop speed (ft)”, “crest curve:”
2 * S – 2158 / A → L
If S > L
Then
Disp L
Else
A * S² / 2158 → L
Disp L
End
Disp “sag curve:”
2 * S – (400 + 3.5 * S) / A → G
If S > G
Then
Disp G
Else
(A * S²) / (400 + 3.5 * S) → G
Disp G
End

Examples:

Example 1:
Grade 1: -1.75%
Grade 2: 2.25%
Design Speed: 40 mph

Result:
Minimum Vertical Curve Length
Crest Curve: 70.5 ft
Sag Curve: 243.125 ft

Example 2:
Grade 1: -1%
Grade 2: 1.7%
Design Speed: 50 mph

Result:
Minimum Vertical Curve Length
Crest Curve: 50.740740741 ft
Sag Curve: 150.925925926 ft

Source:

Michael R. Lindberg, PE “Civil Engineering Reference Manual for the PE Exam” 11th Ed. Professional Publications, Inc: Belmont, CA. 2008. ISBN 13-978-1-59126-192-2

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.


Sunday, May 19, 2013

HP35S Vertical Curve: Elevation at Peak and at End Point


HP35S Vertical Curve: Elevation at Peak and Length

Source: Fundamental of Engineering Supplied-Reference Handbook 8th Ed, 2nd Revision, 2011, NCEES

Calculator: HP 35S

Input
I = initial height
G = grade 1; entering the curve (in decimal form)
H = grade 2; exiting the curve (in decimal form)
L = horizontal length of the curve

G and H are the opposite signs

If G>0 and H<0 a="" and="" curve="" g="" h="" has="" if="" peak.="" similarity="" the="">0, the curve has a valley.

The program listed prompts for the inputs.

Output
1. Point where the curve reaches extreme elevation
2. Press R/S to get the elevation at the extreme point.
3. Press R/S once more to get the elevation at the end of the curve.

None of the outputs are stored.

Formula
A = (G - H)/(2L)
X_extrema = -G/(2A)
Equation of the Curve: y = I + G*x + A*x^2

Examples
Uphill curve:
I = 1,000 ft
G = 7% = 0.07
H = -4% = -0.04
L = 1,368 ft

Point at peak elevation is 870.545 ft into the curve at 1,030.469 ft. The elevation at the end of the curve is 1,020.520 ft.

Downhill curve:
I = 1,580 ft
G = -3% = -0.03
H = 4.2% = 0.042
L = 2.3 mi = 12,144 ft

Point at trough elevation occurs 5,060 ft into the curve at elevation 1,504.1 ft. The elevation at the end of the curve is at 1,652.864 ft.

Program
V001 LBL V
V002 INPUT I
V003 INPUT G
V004 INPUT H
V005 INPUT L
V006 RCL H
V007 RCL - G
V008 2
V009 ÷
V010 RCL ÷ L
V011 STO A
V012 RCL G
V013 x<>y
V014 ÷
V015 2
V016 ÷
V017 +/-
V018 R/S \\ shows point where extreme elevation occurs
V019 XEQ V024
V020 R/S \\ shows extreme elevation
V021 RCL L
V022 XEQ V024 \\ shows ending elevation
V023 RTN
V024 ENTER \\ calculate y subroutine
V025 RCL × A
V026 RCL + G
V027 ×
V028 RCL + I
V029 RTN



This blog is property of Edward Shore. 2013

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