Fortran 90学习之旅(一)Compaq Visual Fortran 6.5 的安装与第一个例子

 

高尔夫球模型的算法是用Fortan 90写的，以前从来没接触过！！！！

从现在开始学习吧。

Fortran语言的最大特性是接近数学公式的自然描述，在计算机里具有很高的执行效率。

易学，语法严谨。

可以直接对矩阵和复数进行运算，这一点类似matlab。

自诞生以来广泛地应用于数值计算领域，积累了大量高效而可靠的源程序。

很多专用的大型数值运算计算机针对Fortran做了优化。

广泛地应用于并行计算和高性能计算领域。

Fortran90，Fortran95，Fortran2003的相继推出使Fortran语言具备了现代高级编程语言的一些特性。

 

Fortran 90 不是软件，是语言规范。

Compaq Visual Fortran 是国内常用的支持 Fortran 90 的编译器.

 

下载地址:

http://dc.whu.edu.cn/bencandy.php?fid=8&id=177

安装成功后的Visual Fortran与VC6.0的界面很像

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由于要在Fotran中要画图，究竟怎么实现呢？

参考 http://hi.baidu.com/desiree100/blog/item/e769a7b78db046f531add19b.html

Visual Fortran中有提供绘图功能。Visual Fortran的绘图功能不完全是以扩充函数的类型存在，使用它的绘图功能必须在选择Project类型时，选择Standard Graphics或QuickWin模式。

Visual Fortran的QuickWin及Standard Graphics模式在简单的绘图使用上会比较方便，它的绘图函数功能比较多样，不过效率会比较差，而且不支持动画功能。

        Standard Graphics和QuickWin模式在绘图方面的使用方法完全相同，它们都是调用相同的函数来绘图。差别在于Standard Graphics只能打开一个窗口来绘图、QuickWin模式则可以打开多个窗口来绘图。QuickWin模式下可以有菜单及对话窗的功能，Standard Graphics则不行。Standard Graphics模式的程序代码可以原封不动直接转换到QuickWin模式下使用，但是QuickWin的程序代码并不一定可以直接拿到Standard Graphics模式下使用。

        我试着运行一个画y = sin(x)的工程。

新建一个工程->选择Fortran Standard Graphics or QuickWin Application。

注意: 工程路径不能含有中文。

新建sinGraph.f90文件

! sin函数的绘图范例program Plot_Sineuse DFLIBimplicit noneinteger, parameter :: lines=500    ! 用多少线段来画函数曲线real(kind=8), parameter :: X_Start=-5.0    ! x轴最小范围real(kind=8), parameter :: X_End=5.0       ! x轴最大范围   real(kind=8), parameter :: Y_Top=2.0       ! y轴最大范围 real(kind=8), parameter :: Y_Bottom=-2.0   ! y轴最小范围integer :: result          ! 取回绘图函数运行状态integer(kind=2) :: color   ! 设定颜色用real(kind=8) :: step       ! 循环的增量real(kind=8) :: x,y        ! 绘图时使用,每条小线段都连接real(kind=8) :: NewX,NewY ! (x,y)及(NewX,NewY)real(kind=8), external :: f ! 待绘图的函数type(wxycoord) :: wt       ! 返回上一次的虚拟坐标位置type(xycoord)   :: t        ! 返回上一次的实际坐标位置! 设定虚拟坐标范围大小    result=SetWindow( .true. , X_Start, Y_Top, X_End, Y_Bottom )! 用索引值的方法来设定颜色result=SetColor(2)    ! 内定的2号是应该是绿色call MoveTo(10,20,t) ! 移动画笔到窗口的(10,20)call OutGText("f(x)=sin(x)")   ! 写出内容! 使用全彩RGB 0-255的256种色阶来设定颜色color=RGBToInteger(255,0,0)        ! 把控制RGB的三个值转换到color中result=SetColorRGB(color)          ! 利用color来设定颜色call MoveTo_W(X_Start,0.0_8,wt)    ! 画X轴result=LineTo_W(X_End,0.0_8)       ! call MoveTo_W(0.0_8,Y_Top,wt)      ! 画Y轴result=LineTo_W(0.0_8,Y_Bottom)    !   step=(X_End-X_Start)/lines         ! 计算小线段间的X间距 ! 参数#FF0000是使用16进制的方法来表示一个整数result=SetColorRGB(#FF0000)         ! 开始绘制小线段们do x=X_Start,X_End-step,step    y=f(x)           ! 线段的左端点    NewX=x+step         NewY=f(NewX)     ! 线段的右端点    call MoveTo_W(x,y,wt)    result=LineTo_W(NewX,NewY)end do! 设定程序结束后,窗口会继续保留result=SetExitQQ(QWIN$EXITPERSIST)end! 所要绘图的函数real(kind=8) function f(x)implicit none  real(kind=8) :: x  f=sin(x)  returnend function f

马上就编译通过了.

运行结果为

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试了一个Demo后现在该上演自己的golfModem代码了.

! Final Project for Fortran - Andrew Werner - March 13, 2007 - Flight of a Golf!Ball! Design a project that allows you to investigate a physics problem in depth!throw the use of a Fortran program.! The problem I chose to investigate is the flight path of a golf ball.! Assumptions being used(made)! *The golf shot is being made at standard ambient temperature and pressure of!25 degrees C (70 F) and 100KPa.! That the average driver club head mass is 200 grams.! The drag coefficient of a golf ball is 0.21! The lift coefficient of a golf ball is 0.14! The radius of a golf ball is 21.335 mm! The mass of a golf ball is 45.93 grams! The coefficient of restitution of the club being used is at the USGA limit of!0.83! The average mass of a driver club head is 200 grams! *The golf ball maintains a constant spin rate throughout the flight of the!shot.! *The launch angle of the shot is 3 degrees more than the loft of the driver.! *The rotation of the golf ball is only in the x-direction and does not have!any (what would be) z-components! That the force of gravity is a constant, would actually depend upon elevation!=> distance from the center of the earth.PROGRAM golf_ballIMPLICIT NONEREAL(KIND=8)::launch_angle,swing_speed,loft_club,ball_speed,delta_t=.001,lC=0.14,dC=0.21,air_densityREAL(KIND=8)::mass_clubhead, drag_force, lift_force, g=9.8,radius,ball_massREAL(KIND=8)::area_cross, swing_speed_mph, launch_angle_deg, last_pointREAL,DIMENSION(100000)::a_drag,a_lift,a_x,v_x,x_m,a_y,v_y,y_m,theta,V,y_yards,x_yardsINTEGER::iINTEGER:: PGBEG, IER, max_distance_m, max_distance_yds!LOGICAL:: y_yREAL(KIND=8)::one=1,zero=0,n_one=(-1),pi!used to determine the value of pipi = (ACOS(n_one))!Launch angle = launch angle of the golf ball in relation to the ground, figure!from loft of club, in degrees!swing_speed = the speed of the particular golfer, prompt for mph then convert!to m/s!loft_club = the loft of the club being used in degrees!ball_speed = the speed of the golf ball in m/s!t = time in seconds!delta_t = the increment of time, being used to create a number of heights at!certain t's!lC = Lift coefficient!dC = drag coefficient!air_density = density of the air being traveled through, in g/m^3!radius = golf ball radius in mm!ball_mass = mass of the golf ball in grams!mass_clubhead = mass of the club head of the golf club!area_cross = the cross sectional area of a golf ball, pi*r^2!drag_force = the drag force felt by the golf ball!lift_force = the lift force felt by the golf ballPRINT *, 'Welcome to the driver distance figure-outer'PRINt *, ' The driver distance figure outer will ask you for your swing speedalong with the loft of your driver,'PRINT *, ' and in return figure out how far your shot will travel.'PRINT *, "So let's get started."PRINT *,' 'PRINT *, 'What is your swing speed in mph?'READ *, swing_speed_mphPRINT *, ' 'PRINT *, 'What is the loft of you driver (in degrees)?'READ *, loft_club!To decrease complications in this project, I will use the case of standard!temperature and pressure (25 C and 100KPA)!At standard ambient temperature and pressure the density of air is 1.168!kg/(m^3 => 1168 g/(m^3)air_density = 1.168 !kg/m^3!The properties of the golf ball of radius and mass will be those as defined by!the USGA in the rules of golf.!The minimum diameter of a golf ball is specified as 42.67 mm, radius =>21.335!mm.radius = .021335 !m!The maximum mass of the golf ball is specified as 45.93 grams by the USGA in!the rules of golf.ball_mass = .04593 !kg!The average mass of the driver club head is said to be 200 grams.mass_clubhead = .200 !kg!Here I am going to convert the ball_speed_mph from mph into meters/second!5280 ft = 1 mile, 3600 secs = 1 hour, .304800610 m = 1 ftswing_speed = swing_speed_mph * (5280./3600.) * .30480061!Here is the relationship between the swing speed and the resulting ball speed.!By the conservation of momentum, we know m1v1 = m2v2!ball_speed = (swing_speed * ball_mass) / clubhead_mass This equation would!seem to be of the correct form,!However one of the hot topics in golf is the coefficient restitution,!which is a measure of the efficiency of the kinetic energy transfer between!club and ball! The USGA has also put a limit on this attribute of the golf club, which is!0.83! Applying this attribute to our conservation of momentum equation we getball_speed = swing_speed * ( (1.+0.83)/(1.0+ (ball_mass/mass_clubhead) ) )!For the launch angle, I'm going to take into consideration that most golfers!catch the ball on the slight! upswing with their driver, adding a few degrees on top of their club head loft!for their launch anglelaunch_angle_deg = loft_club + 3.0launch_angle = (launch_angle_deg / 180) * pi !Converting the launch angle into radians in order to use intrinsic functions.!In order to be congruent with the forms of equations I am using from my!derivations I will define! the launch angle to = theta, and the angle of (90 - theta) to = phi!theta = launch_angle!phi = 90. - theta!For the cross_sectional area of the golf ball, we get the form ofarea_cross = pi * (radius**2.)!For the value of pi, above I used a method from one of my previous homeworks!Now that we have the initial velocity of the ball and the launch angle, we can!go to town on the! equations that determine the flight of the ball! The forces on the golf ball are the drag force, the lift force, and the force!of gravity.! While the force of gravity is always downward (-y), the drag force is always!parallel to the direction of motion,! while the lift force is always perpendicular to the direction of motion, and!in the case of a golf ball,! since there is always backspin imparted on the golf ball the lift force will!always be upward (+y)!From these thoughts we can determine that the both the lift force and the drag!force will have horizontal and!vertical components.!Using the form of F=ma, and Ftotal = Fdrag + Flift + mg = ma, we can come up with the forms of! a = (Flift + Fdrag + mg) / ball_mass!Separating this form of the acceleration of the golf ball into x and y!components yields! x direction: a_x = ( lift_force*COS(phi) + drag_force*SIN(theta) ) / ball_mass! y direction: a_y = ( lift_force*SIN(phi) + drag_force*COS(theta) + ball_mass*g!) / ball_mass! After taking two derivatives of y''=a, y'=at + v, y=(1/2)at^2 + vt + y, and!the same for x, I come up with! x = .5*a_x*(t**2) + t*(ball_speed*SIN(theta))! y = .5*a_y*(t**2) + t*(ball_speed*COS(theta))!As equations of the y and x ball positions as a function of time (and currently!theta)!Now for the drag and lift forces.!lift_force = .5*air_density*(ball_speed**2)*cross_area*lC!drag_force = .5*air_density*(ball_speed**2)*cross_area*dC!lift_force = .5*air_density*(ball_speed**2)*area_cross*lC!drag_force = .5*air_density*(ball_speed**2)*area_cross*dC!a_x = ( -lift_force*COS(phi) - drag_force*SIN(theta) ) / ball_mass!a_y = ( lift_force*SIN(phi) - drag_force*COS(theta) - ball_mass*g ) / ball_masstheta(1) = launch_angle!With V(i) being the magnitude of the velocity of the golf ball, we can us V = v_x^2 + v_y^2 for the magnitudev_x(1) = ball_speed * COS(launch_angle)v_y(1) = ball_speed * SIN(launch_angle)V(1) = SQRT( v_x(1)**2. + v_y(1)**2. )a_x(1) = 0a_y(1) = 0x_m(1) = 0y_m(1) = 0a_lift(1) = 0a_drag(1) = 0!PRINT *, swing_speed, ball_speed!PRINT *, a_y(1:5)!PRINT *, v_y(1:5)!PRINT *, y(1:5)!PRINT *,' '!PRINT *, air_density, area_cross, dC, ball_mass, lC, V(1)DO i=2,100000a_drag(i) = ( (.5*air_density*area_cross*dC)/ball_mass) * (V(i-1)**2.)a_lift(i) = ( (.5*air_density*area_cross*lC)/ball_mass) * (V(i-1)**2.)a_x(i) = a_drag(i)*(-COS(theta(i))) + a_lift(i)*(-SIN(theta(i-1)))v_x(i) = v_x(i-1) + a_x(i)*delta_tx_m(i) = x_m(i-1) + v_x(I)*delta_tx_yards(i) = x_m(i)*(1./0.9144)a_y(i) = -g + a_drag(i)*(-SIN(theta(i))) + a_lift(i)*(COS(theta(i-1)))v_y(i) = v_y(i-1) + a_y(i)*delta_ty_m(i) = y_m(i-1) + v_y(I)*delta_ty_yards(i) = y_m(i)*(1./0.9144) !There are 0.9144 meters in a yard.V(i) = SQRT( v_x(i)**2. + v_y(i)**2. )theta(i) = ATAN( v_y(i) / v_x(i) )last_point = 0IF ( (y_m(i)<0) .AND. (last_point == 0)) THENmax_distance_m = x_m(i)EXITEND IFEND DO!PRINT *, a_drag(1:5) Used for debugging the program.!PRINT *, a_lift(1:5)!PRINT *, a_x(1:5)!PRINT *, v_x(1:5)!PRINT *, x(1:5)!IER = PGBEG(0,'ballflight.ps/PS',1,1)IER = PGBEG(0,'/xserve',1,1) !Naming the graph,选择graphics设备 IF (IER.NE.1) STOPCALL PGENV(0,MAXVAL(REAL(x_yards(1:100000))),0,2*MAXVAL(REAL(y_yards(1:10000))),0,1)!根据最大最小的数据组定义plot的scalesCALL PGLAB('Distance (yds)','Height (yds)','Flight Path of a Golf Ball')!Creating labels for the graphCALL PGPT(20000, REAL(x_yards(1:20000)), REAL(y_yards(1:20000)),20)!在graph中创建点标记CALL PGENDmax_distance_yds = max_distance_m * (1./ 0.9144)!There are 0.9144 meters in a yard.PRINT *,' 'PRINT *, 'The length of you shot was', max_distance_yds,'yards'PRINT *, 'Wow, great shot!'PRINT *, ' 'PRINT *, 'Have a great day!'END PROGRAM golf_ball

编译后提示如下错误:

Linking...GolfModem.obj : error LNK2001: unresolved external symbol [e-mail] : error LNK2001: unresolved external symbol [e-mail] : error LNK2001: unresolved external symbol [e-mail] : error LNK2001: unresolved external symbol [e-mail] : error LNK2001: unresolved external symbol _PGEND@0Debug/GolfModem.exe : fatal error LNK1120: 5 unresolved externalsError executing link.exe.

这是因为代码里的红色部分。它们用OPENGL库绘图，我没有OpenGL库，接下来准备改Visual Fortran自带的Visual Fortran的QuickWin及Standard Graphics来画图了。

以下是修改后的代码:红色为我添加的内容。

! Final Project for Fortran - Andrew Werner - March 13, 2007 - Flight of a Golf!Ball! Design a project that allows you to investigate a physics problem in depth!throw the use of a Fortran program.! The problem I chose to investigate is the flight path of a golf ball.! Assumptions being used(made)! *The golf shot is being made at standard ambient temperature and pressure of!25 degrees C (70 F) and 100KPa.! That the average driver club head mass is 200 grams.! The drag coefficient of a golf ball is 0.21! The lift coefficient of a golf ball is 0.14! The radius of a golf ball is 21.335 mm! The mass of a golf ball is 45.93 grams! The coefficient of restitution of the club being used is at the USGA limit of!0.83! The average mass of a driver club head is 200 grams! *The golf ball maintains a constant spin rate throughout the flight of the!shot.! *The launch angle of the shot is 3 degrees more than the loft of the driver.! *The rotation of the golf ball is only in the x-direction and does not have!any (what would be) z-components! That the force of gravity is a constant, would actually depend upon elevation!=> distance from the center of the earth.PROGRAM golf_balluse DFLIBIMPLICIT NONEtype(wxycoord) :: wt       ! 返回上一次的虚拟坐标位置real(kind=8) :: x,y        ! 绘图时使用,每条小线段都连接real(kind=8) :: NewX,NewY ! (x,y)及(NewX,NewY)integer :: result          ! 取回绘图函数运行状态real(kind=8), parameter :: X_Start=0   ! x轴最小范围real(kind=8), parameter :: X_End=400       ! x轴最大范围   real(kind=8), parameter :: Y_Top=100       ! y轴最大范围 real(kind=8), parameter :: Y_Bottom=0   ! y轴最小范围REAL(KIND=8)::launch_angle,swing_speed,loft_club,ball_speed,delta_t=.001,lC=0.14,dC=0.21,air_densityREAL(KIND=8)::mass_clubhead, drag_force, lift_force, g=9.8,radius,ball_massREAL(KIND=8)::area_cross, swing_speed_mph, launch_angle_deg, last_pointREAL,DIMENSION(100000)::a_drag,a_lift,a_x,v_x,x_m,a_y,v_y,y_m,theta,V,y_yards,x_yardsINTEGER::iINTEGER:: PGBEG, IER, max_distance_m, max_distance_yds!LOGICAL:: y_yREAL(KIND=8)::one=1,zero=0,n_one=(-1),pi!used to determine the value of pipi = (ACOS(n_one))!Launch angle = launch angle of the golf ball in relation to the ground, figure!from loft of club, in degrees!swing_speed = the speed of the particular golfer, prompt for mph then convert!to m/s!loft_club = the loft of the club being used in degrees!ball_speed = the speed of the golf ball in m/s!t = time in seconds!delta_t = the increment of time, being used to create a number of heights at!certain t's!lC = Lift coefficient!dC = drag coefficient!air_density = density of the air being traveled through, in g/m^3!radius = golf ball radius in mm!ball_mass = mass of the golf ball in grams!mass_clubhead = mass of the club head of the golf club!area_cross = the cross sectional area of a golf ball, pi*r^2!drag_force = the drag force felt by the golf ball!lift_force = the lift force felt by the golf ballPRINT *, 'Welcome to the driver distance figure-outer'PRINt *, ' The driver distance figure outer will ask you for your swing speedalong with the loft of your driver,'PRINT *, ' and in return figure out how far your shot will travel.'PRINT *, "So let's get started."PRINT *,' 'PRINT *, 'What is your swing speed in mph?'READ *, swing_speed_mphPRINT *, ' 'PRINT *, 'What is the loft of you driver (in degrees)?'READ *, loft_club!To decrease complications in this project, I will use the case of standard!temperature and pressure (25 C and 100KPA)!At standard ambient temperature and pressure the density of air is 1.168!kg/(m^3 => 1168 g/(m^3)air_density = 1.168 !kg/m^3!The properties of the golf ball of radius and mass will be those as defined by!the USGA in the rules of golf.!The minimum diameter of a golf ball is specified as 42.67 mm, radius =>21.335!mm.radius = .021335 !m!The maximum mass of the golf ball is specified as 45.93 grams by the USGA in!the rules of golf.ball_mass = .04593 !kg!The average mass of the driver club head is said to be 200 grams.mass_clubhead = .200 !kg!Here I am going to convert the ball_speed_mph from mph into meters/second!5280 ft = 1 mile, 3600 secs = 1 hour, .304800610 m = 1 ftswing_speed = swing_speed_mph * (5280./3600.) * .30480061!Here is the relationship between the swing speed and the resulting ball speed.!By the conservation of momentum, we know m1v1 = m2v2!ball_speed = (swing_speed * ball_mass) / clubhead_mass This equation would!seem to be of the correct form,!However one of the hot topics in golf is the coefficient restitution,!which is a measure of the efficiency of the kinetic energy transfer between!club and ball! The USGA has also put a limit on this attribute of the golf club, which is!0.83! Applying this attribute to our conservation of momentum equation we getball_speed = swing_speed * ( (1.+0.83)/(1.0+ (ball_mass/mass_clubhead) ) )!For the launch angle, I'm going to take into consideration that most golfers!catch the ball on the slight! upswing with their driver, adding a few degrees on top of their club head loft!for their launch anglelaunch_angle_deg = loft_club + 3.0launch_angle = (launch_angle_deg / 180) * pi !Converting the launch angle into radians in order to use intrinsic functions.!In order to be congruent with the forms of equations I am using from my!derivations I will define! the launch angle to = theta, and the angle of (90 - theta) to = phi!theta = launch_angle!phi = 90. - theta!For the cross_sectional area of the golf ball, we get the form ofarea_cross = pi * (radius**2.)!For the value of pi, above I used a method from one of my previous homeworks!Now that we have the initial velocity of the ball and the launch angle, we can!go to town on the! equations that determine the flight of the ball! The forces on the golf ball are the drag force, the lift force, and the force!of gravity.! While the force of gravity is always downward (-y), the drag force is always!parallel to the direction of motion,! while the lift force is always perpendicular to the direction of motion, and!in the case of a golf ball,! since there is always backspin imparted on the golf ball the lift force will!always be upward (+y)!From these thoughts we can determine that the both the lift force and the drag!force will have horizontal and!vertical components.!Using the form of F=ma, and Ftotal = Fdrag + Flift + mg = ma, we can come up with the forms of! a = (Flift + Fdrag + mg) / ball_mass!Separating this form of the acceleration of the golf ball into x and y!components yields! x direction: a_x = ( lift_force*COS(phi) + drag_force*SIN(theta) ) / ball_mass! y direction: a_y = ( lift_force*SIN(phi) + drag_force*COS(theta) + ball_mass*g!) / ball_mass! After taking two derivatives of y''=a, y'=at + v, y=(1/2)at^2 + vt + y, and!the same for x, I come up with! x = .5*a_x*(t**2) + t*(ball_speed*SIN(theta))! y = .5*a_y*(t**2) + t*(ball_speed*COS(theta))!As equations of the y and x ball positions as a function of time (and currently!theta)!Now for the drag and lift forces.!lift_force = .5*air_density*(ball_speed**2)*cross_area*lC!drag_force = .5*air_density*(ball_speed**2)*cross_area*dC!lift_force = .5*air_density*(ball_speed**2)*area_cross*lC!drag_force = .5*air_density*(ball_speed**2)*area_cross*dC!a_x = ( -lift_force*COS(phi) - drag_force*SIN(theta) ) / ball_mass!a_y = ( lift_force*SIN(phi) - drag_force*COS(theta) - ball_mass*g ) / ball_masstheta(1) = launch_angle!With V(i) being the magnitude of the velocity of the golf ball, we can us V = v_x^2 + v_y^2 for the magnitudev_x(1) = ball_speed * COS(launch_angle)v_y(1) = ball_speed * SIN(launch_angle)V(1) = SQRT( v_x(1)**2. + v_y(1)**2. )a_x(1) = 0a_y(1) = 0x_m(1) = 0y_m(1) = 0a_lift(1) = 0a_drag(1) = 0!PRINT *, swing_speed, ball_speed!PRINT *, a_y(1:5)!PRINT *, v_y(1:5)!PRINT *, y(1:5)!PRINT *,' '!PRINT *, air_density, area_cross, dC, ball_mass, lC, V(1)! 设定虚拟坐标范围大小    result=SetWindow( .true. , X_Start, Y_Bottom, X_End, Y_Top )call MoveTo_W(X_Start,0.0_8,wt)    ! 画X轴result=LineTo_W(X_End,0.0_8)       ! call MoveTo_W(0.0_8,Y_Top,wt)      ! 画Y轴result=LineTo_W(0.0_8,Y_Bottom)    !  DO i=2,100000a_drag(i) = ( (.5*air_density*area_cross*dC)/ball_mass) * (V(i-1)**2.)a_lift(i) = ( (.5*air_density*area_cross*lC)/ball_mass) * (V(i-1)**2.)a_x(i) = a_drag(i)*(-COS(theta(i))) + a_lift(i)*(-SIN(theta(i-1)))v_x(i) = v_x(i-1) + a_x(i)*delta_tx_m(i) = x_m(i-1) + v_x(I)*delta_tx_yards(i) = x_m(i)*(1./0.9144)a_y(i) = -g + a_drag(i)*(-SIN(theta(i))) + a_lift(i)*(COS(theta(i-1)))v_y(i) = v_y(i-1) + a_y(i)*delta_ty_m(i) = y_m(i-1) + v_y(I)*delta_ty_yards(i) = y_m(i)*(1./0.9144) !There are 0.9144 meters in a yard.V(i) = SQRT( v_x(i)**2. + v_y(i)**2. )theta(i) = ATAN( v_y(i) / v_x(i) )last_point = 0IF ( (y_m(i)<0) .AND. (last_point == 0)) THENmax_distance_m = x_m(i)EXITEND IFx = x_yards(i-1)y = y_yards(i-1)NewX = x_yards(i)NewY = y_yards(i)!PRINT *, 'now x is ', x,'y is',ycall MoveTo_W(x,y,wt)result=LineTo_W(NewX,NewY) END DO!PRINT *, a_drag(1:5) Used for debugging the program.!PRINT *, a_lift(1:5)!PRINT *, a_x(1:5)!PRINT *, v_x(1:5)!PRINT *, x(1:5)!IER = PGBEG(0,'ballflight.ps/PS',1,1)max_distance_yds = max_distance_m * (1./ 0.9144)!There are 0.9144 meters in a yard.PRINT *,' 'PRINT *, 'The length of you shot was', max_distance_yds,'yards'PRINT *, 'Wow, great shot!'PRINT *, ' 'PRINT *, 'Have a great day!'! 设定程序结束后,窗口会继续保留result=SetExitQQ(QWIN$EXITPERSIST)END PROGRAM golf_ball