Question

HW 5-16-submission format protected View) Microsoft Word (Product age Layout References Mailings Review View Add-Ins nated from an Internet location and might be unsafe. Click for more details. Enable E CSCI 1205-001, Spring 2016 HW #7 Functions and Arrays Name: Dr. Gray Due date: 4-10-16 Output: Required-Output Screen C: n1.exe Time: 19.954s X-Pos: 998.613m Y-Pos: 0.815m Height: 0.881 Time 19.955s X-Pos 998.663m Y-Pos: -0.812m Height: 0.783 Time: 19.956s X-Pos 998.713m Y-Pos: -0.810m Height: e.685 Time 19.957s X-Pos 998.763a V-Pos: -0.807m Height: 0.587 Time 19.958s X-Pos. 998.813m Y-Pos: 0.805m Height: 0.489 Time 19.959s X-Pos: 998.863 Y-pos 0.803m Height: 0.391 Time 19.960s X-Pos: 998. 914m Y-Pos: 0.800m Height: 0.293 Time 19.961 s X-Pos: 998.964m Y-Pos 0.798 m Height: 0.196 Time 19.962s X-Pos: 999.014m Y-Pos: 0.795m Height: 0.098 Time 19.963s X-Pos. 999 .064n Y-Pos: 0.793m Height: 0.000 missed by 1.227 Author: Dr. Gray Azimuth -2.900 Elevation angle 62.900 Program Listing: Your code goes here Pseudocode listing:

HW#5 Functions and Arrays

For HW#5 we will return to the state method of solving the ballistic projectile problem with added twists. The goal will be to hit a target 1000 meters east of your cannon but the wind is blowing from the south. Your initial muzzle velocity is fixed at 110 meters per second, but you can control the elevation and azimuth angles of your aim. Elevation angle is measured up or down. Azimuth angle is measured right or left. Use positive x for east, positive y for north, and positive z for up.

For HW#5 you must use an iterative solution with a time increment of .001 seconds. Even if you could derive an exact solution for the position of the projectile at any instant in time, you may not use it.

You must use arrays to store the x, y, and z components of position, velocity, and acceleration. You must use functions to update the position and velocity arrays and output the current x, y, z position of the projectile after each time increment. You will terminate the iteration when the projectile hits the ground and calculate the horizontal distance from the impact point to the actual target. The closest shot wins bonus points.

Your working variables will be limited to a constant time-increment, aiming inputs and three arrays; position, velocity, and acceleration. Each array will contain 3 elements representing x, y, and z components.

Function prototypes

Void UpdatePosistion(double pos[], double vel[], const double timeInc);

Void UpdateVelocity(double vel[], const double acc[], const double timeInc);

Void OutputPosition(double pos[]);

Variables

const double timeInc = .001 // unit s

const double initialVelocity = 110; // units m/s

const double target[3] = { 1000, 0, 0}; // units m ( x, y, z coordinates) East is positive x

const double acc[3] = { 0, .25, -9.81}; // units m/s² wind and gravity, North is positive y

double azimuth, elevation, time;

double pos[3], vel[3];

As always you must match the column and precision formatting.

Don’t for get to sign your screens.

HW#5 Ballistics

State Machine Calculation of Trajectories

The objective of homework 5 is to calculate and display the ballistic flight trajectory of a projectile. The problem is simple, an artillery shell is fired from a cannon with an initial velocity, azimuth, and an elevation angle. This produces an initial velocity vector with 3-dimensional components. Over time the force of gravity causes the initial upward velocity to decrease and eventually become downward. From that point on the projectile accelerates downward until it strikes the ground. The time it takes to hit the ground is called the flight time. During the flight time the projectile also has a horizontal component of velocity which cause it to travel downrange. In homework 5 there is also a variable headwind in the horizontal direction which slows the horizontal velocity over time.

In a simple problem there are no horizontal forces acting on the projectile and the entire problem could easily be solved by a simple kinematic equation which could yield exact height and range positions at any instant of time. The addition of a variable headwind significantly complicates the problem and makes ballistic equations difficult to use and certainly beyond the scope of a freshmen class. Fortunately there is another option called a “state machine”.

At any instant of time the projectile has a range and height position, vertical and horizontal components of velocity and vertical and horizontal components of acceleration. Its’ state is completely described by 6 variables. In any small interval of time the change from one state to the next state can be calculated by simple equations. If the time interval between states is small enough the accuracy of this approximate method is quite high.

Equations

newXpos = oldXpos + timeInterval * oldXVel

newYpos = oldYpos + timeInterval * oldYVel

newXvel = oldXvel + timeInterval * Xacceleration

newYvel = oldYvel + timeInterval * Yacceleration

Yacceleration = -9.81 + time

X acceleration = -time // it changes with time

muzzleVelocity = 50;

InitialXVel = muzzleVelocity * cos(elevationAngle)

InitialYvel = muzzleVelocity * sin(elevationAngle)

timeInterval = .001 seconds

Units

Seconds

Meters

Course objectives

You must use functions to update position, velocity, and acceleration.

You must use arrays for the position, velocity, and acceleration vectors.

Bonus points for students who can hit the target.

You must prompt the user for initial input of azimuth and elevation angle.

You must output the time, the height, and the downrange distance at each increment of time. (instantaneous states)

You program must terminate when the projectile strikes the ground.

You must use the “+=” operator.

All data must be output with three decimal places. (use setprecision)

All columns must be aligned. (use setw)

Answer 1

Executable code :

#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;

void UpdatePosition(double pos[], double vel[], const double timeInc);
void UpdateVelocity(double vel[], const double acc[], const double timeInc);
void OutputPosition(double pos[]);

int main()
{
const double PI=3.142;
const double timeInc = .001;
const double initialVelocity = 110;
const double target[3] = { 1000, 0, 0};
const double acc[3] = { 0, .25, -9.81};
double azimuth, elevation, time;
double pos[3], vel[3];

for(int i=0;i<3;i++)
{
pos[i]=0;
}

cin>>elevation;
cout<<"enter the azimuth angle"<<endl;
cin>>azimuth;

  
vel[0]=initialVelocity*cos(elevation*PI/180)*cos(azimuth*PI/180);
vel[1]=initialVelocity*cos(elevation*PI/180)*sin(azimuth*PI/180);
vel[2]=initialVelocity*sin(elevation*PI/180);

double flightTime=0.0;
while(pos[2]>0 || flightTime==0)
{
flightTime+=timeInc;
UpdatePosition(pos,vel,timeInc);
UpdateVelocity(vel,acc,timeInc);
cout<<"Time "<<setw(10)<<setprecision(6)<<flightTime<<"s ";
OutputPosition(pos);
}

cout<<"missed by "<<setprecision(4)<<abs(1000-pos[0])<<endl;
cout<<"Azimuth = "<<setprecision(4)<<azimuth<<" Elevation = "<<setprecision(4)<<elevation<<endl;

}

void UpdatePosition(double pos[], double vel[], const double timeInc)
{
pos[0]+=+timeInc*vel[0];
pos[1]+=+timeInc*vel[1];
pos[2]+=timeInc*vel[2];
}
void UpdateVelocity(double vel[], const double acc[], const double timeInc)
{

vel[1]+=timeInc*acc[1];
vel[2]+=timeInc*acc[2];
}
void OutputPosition(double pos[])
{
cout<<"X-POS "<<setw(10)<<setprecision(6)<<pos[0]<<"m ";
cout<<"Y_POS"<<setw(10)<<setprecision(3)<<pos[1]<<"m ";
cout<<"Height"<<setw(10)<<setprecision(3)<<pos[2]<<"m"<<endl;
}

#include #include #include iostream» <cmath> doman!p> 4 using namespace std; 7 void UpdatePosition(double pos[l, double vel[l, const double timeInc); 8 void UpdateVelocity (double vel[1, const double acc[], const double timeInc); 9 void OutputPosition(double pos[]); 10 11 int main() 12 13 14 15 16 const double PI-3.142; const double timeInc- .001; const double initialVelocity - 110; const double target[3]- 1000, e, 0 const double acc3] 1 0, .25, -9.81; double azimuth, elevation, time; double pos [3], vel[3]; 18 19 20 21 for( int i-e;i< 3; i++) 23 pos[i]-0; 25 26 cin>>elevation; cout < < "enter the cin>>azimuth; azimuth angle"(«endl; 28 29 31 32 vel[0]-initialVelocity cos (elevation*PI/180)*cos (azimuth*PI/180); vel[1]-initialVelocity*cos (elevation*PI/180) sin(azimuth*PI/180); vel[2]-initialVelocity sin(elevation*PI/180); 34 double flightTime-0.0; 36 while(pos[2]>0 II flightTime-0) 40 41 42 43 flightTime+=time!nc; UpdatePosition(pos,vel,timeInc);l UpdateVelocity (vel,acc,timeInc); cout<<"Time "<<setw (10)<<setprecision(6)<<flightTime<<"S " OutputPosition(pos); 45 46 Get URL options compilation execution enter azimtuh enter the azimuth angle Time missed by 999.9 Azimuth4.941e-324 Elevation0 0.001s X-POS 0.11m Y POS Om Height 0m