Пример курсовой работы по Delphi (Кафедра)

61

Продолжение табл. 5.1

1	2	3	4
118	Transparent	False	TImage(Image1)
119	Visible	True
120	Width	409
121	Align	alTop
122	Anchors	[akLeft,akTop,akRight]
122.1	akLeft	True False
122.2	akTop	True
122.3	akRight	True
122.4	akBottom	False
123	AutoSize	False
124	BorderWidth	0
125	ButtonHeight	22
126	ButtonWidth	24
127	Caption	ToolBar1
128	Color	$00C9DDC8
129	Constraints	(TSizeConstraints)
129.1	MaxHeight	0
129.2	MaxWidth	0
129.3	MinHeight	0
129.4	MinWidth	0
130	Cursor	crDefault
131	DisabledImages
132	DockSite	False
133	DragCursor	crDrag
134	DragKind	dkDrag	TToolBar(ToolBar1)
135	DragMode	dmManual
136	EdgeBorders	[ebTop]
136.1	ebLeft	False
136.2	ebTop	True
136.3	ebRight	False
136.4	ebBottom	False
137	EdgeInner	esRaised
138	EdgeIOuter	esLowered
139	Enabled	True
140	Flat	False
141	Font	(TFont)
141.1	Charset	DEFAULT_CHARSET
141.2	Color	clWindowText
141.3	Height	-11
141.4	Name	MS Sans Serif
141.5	Pitch	fpDefault
141.6	Size	8
142	Height	29
143	HelpContext	0
144	Hint
145	HotImages
146	Images	ImageList1
147	Indent	0

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62

Продолжение табл. 5.1

	1	2	3	4
	148	Left	0
	149	List	False
	150	Name	ToolBar1
	151	ParentColor	True
	152	ParentFont	True
	153	ParentShowHint	True
	154	PopupMenu
	155	ShowCaptions	False	TToolBar(ToolBar1)
	156	ShowHint	False
	157	TabOrder	0
	158	TabStop	False
	159	Tag	0
	160	Top	0
	161	Transparent	False
	162	Visible	True
	163	Width	613
	164	Wrapable	True
	165	Action
	166	AllowAllUp	False
	167	AutoSize	False
	168	Caption	ToolButton6
	169	Cursor	crDefault
	170	Down	False
	171	DragCursor	crDrag
	172	DragKind	dkDrag
	173	DragMode	dmManual
	174	DropdownMenu
	175	Enabled	True
	176	Grouped	False
	177	Height	22
	178	Hint	Создать новые исходные
			данные	TtoolButton
	179	ImageIndex	5
				(ToolButton6)
	180	Indeterminate	False
	181	Left	0
	182	Marked	False
	183	MenuItem
	184	Name	ToolButton6
	185	ParentShowHint	False
	186	PopupMenu
	187	ShowHint	True
	188	Style	tbsButton
	189	Tag	0
	190	Top	2
	191	Visible	True
	192	Width	24
	193	Wrap	False

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63

Продолжение табл.5.1

1	2	3	4
194	DragCursor	crDrag
195	DragKind	dkDrag
196	DragMode	dmManual
197	Enabled	True
198	Font	(TFont)
198.1	Charset	DEFAULT_CHARSET
198.2	Color	clWindowText
198.3	Height	-16
198.4	Name	MS Sans Serif
198.5	Pitch	fpDefault
198.6	Size	12
199	Height	28
200
201
202	Hint		Tedit(Edit_x1)
203	Layout	tlTop
204	Left	8
205	Name	Label1
206	ParentBiDiMode	True
207	ParentColor	True
208	ParentFont	False
209	ParentShowHint	True
210	PopupMenu
211	ShowAccelChar	True
212	ShowHint	False
213	Tag	0
214	Top	40
215	Transparent	False
216	Visible	True
217	Width	171
218	WordWrap	True

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64

ПРИЛОЖЕНИЕ Е -ТЕСТОВЫЕ ПРИМЕРЫ В MATHCAD

Пример 1

Тестовый пример 1 для

апробации программного продукта Задача: вычисление

прямоугольных координат пунктов геодезической

сети, определяемых обратн засечкой (Ю.Рандярв)

b1 := ( 102 39 37)	b2 := ( 54 47 41)
XA := 8429.93	YA := 6016.34
XB := 6273.45	YB:= 4008.30
XC := 7373.14	YC:= 2737.12
XD := 10491.83	YD:= 3824.62

éæ

b1

, 1

b1

, 2

ö

p

ù

b1 := cotêçb1

, 0

+

0

+

0

÷

×

ú

60

3600

ëè

0

ø

180û

éæ

b2

, 1

b2

, 2

ö

p

ù

b2 := cotêçb2

, 0

+

0

+

0

÷

×

ú

60

3600

ëè

0

ø

180û

b3 := ( 110 24 48)

b1 = -0.224631

b2 = 0.705560

X1 := XA + (YA - YB) × b1				Y1:= YA + ( XB - XA) × b1
X2 := XC + (YB- YC) × b2				Y2:= YC+ ( XC - XB) × b2
X2 = 8270.034				Y2 = 3513.018
Контроль правильности вычислений
XB_XA :=			[b1 × (Y1 - YB) - ( X1 - XB)]				XB_XA = -2156.480	XB - XA = -2156.480
			1 + b12
XC_XB:=		[b2 × (Y2 - YB) + ( X2 - XB)]					XC_XB = 1099.690	XC - XB = 1099.690
			1 + b22
YA_YB:=	[b1 × ( X1 - XB) + (Y1 - YB)]						YA_YB= 2008.040	YA - YB = 2008.040
			1 + b12

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65

YB_YC:=	[b2 × ( X2 - XB) - (Y2 - YB)]							YB_YC= 1271.180			YB - YC = 1271.180
			1 + b22
Вычисляем дирекционный угол ВМ
tg_BM := X2 - X1					tg_BM = 0.097456
	Y1 - Y2
ctg_BM :=		Y1 - Y2			ctg_BM = 10.261033 a := tg_BM					c := ctg_BM
		X2 - X1
Определяем координаты точки М
S1 := a × (Y1 - YB) + ( X1 - XB)						S1 = 1948.316
S2 := a × (Y2 - YB) + ( X2 - XB)						S2 = 1948.316
XM := XB +			S1		XM = 8203.436			XM := XB +	S2		XM = 8203.436
			1 + a2						1 + a2
YM := YB + a × ( XM - XB) YM = 4196.389
Контроль
S3 := c × ( X1 - XB) + (Y1 - YB)							S3 = 19991.734
S4 := c × ( X2 - XB) + (Y2 - YB)							S4 = 19991.734
XM := XB + c × (YM - YB)						XM = 8203.436
YM := YB + (YM - YB)						YM = 4196.389
Заключительный контроль
b_MD := dir_ug( XM,YM, XD,YD)								b_MD = 6.122			ugol(b_MD) = ( 350 46 21)
b_MC := dir_ug( XM,YM, XC, YC)								b_MC = 4.195
											ugol(b_MC) = ( 240 21 40)
b3_k := b_MD - b_MC
bk_3 := ugol(b3_k)					bk_3 = ( 110			24 41)

Пользовательские подпрограммы

ugol(ug) º ug ¬ ug ×		180		dir_ug(x1,y1, x2,y2) º
		p
gr ¬ trunc(ug)
min ¬ trunc[(ug - gr) × 60]
sec ¬ round[[(ug - gr) × 60 - min] × 60,1]
bk30, 0	¬ gr
bk30, 1	¬ min
bk30, 2	¬ sec

return bk3

dx ¬ x2 - x1

dy ¬ y2 - y1

if

dx

0

ug ¬ p

if

dy > 0

2

ug ¬ 0

if

dy

0

ug ¬ p × 3

if

dy < 0

2

if

dx > 0

ug1 ¬ atan

æ dy ö

è dx ø

ug ¬ ug1

if

dy ³ 0

ug ¬ 2 × p -

ug1

if dy < 0

if

dx < 0

ug1 ¬ atan

æ dy ö

è dx ø

ug ¬ p -

ug1

if

dy ³ 0

ug ¬ p +

ug1

if

dy < 0

return

ug

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66

Пример 2

Тестовый пример 2 для

апробации программного продукта Задача: вычисление

прямоугольных координат пунктов геодезической

сети, определяемых обратн засечкой (Ю.Рандярв)

b1 := ( 25 20 50)	b2 := ( 36 01 40)	b3 := ( 35 26 39)
XA := 6789.402	YA := 10786.515
XB := 7595.886	YB:= 10495.656
XC := 9255.486	YC:= 11044.262
XD := 8965.634	YD:= 12666.468

éæ

b1

, 1

b1

, 2

ö

p

ù

b1 := cotêçb1

, 0

+

0

+

0

÷

×

ú

b1 = 2.111011

60

3600

ëè

0

ø

180û

éæ

b2

, 1

b2

, 2

ö

p

ù

b2 := cotêçb2

, 0

+

0

+

0

÷

×

ú

b2 = 1.374980

60

3600

ëè

0

ø

180û

X1 := XA + (YA - YB) × b1					Y1:= YA + ( XB - XA) × b1
X1 = 7403.409					Y1 = 12489.012
X2 := XC + (YB - YC) × b2					Y2:= YC + ( XC - XB) × b2
X2 = 8501.164					Y2 = 13326.178
Контроль правильности вычислений
XB_XA :=				[b1 × (Y1 - YB) - ( X1 - XB)]					XB_XA = 806.484	XB - XA = 806.484
				1 + b12
XC_XB:=			[b2 × (Y2 - YB) + ( X2 - XB)]						XC_XB= 1659.600	XC - XB = 1659.600
				1 + b22
YA_YB:=		[b1 × ( X1 - XB) + (Y1 - YB)]							YA_YB= 290.859	YA - YB= 290.859
				1 + b12
YB_YC:=	[b2 × ( X2 - XB) - (Y2 - YB)]								YB_YC= -548.606	YB- YC = -548.606
2

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67

Вычисляем дирекционный угол ВМ
tg_BM := X2 - X1				tg_BM = -1.311275
Y1 - Y2
ctg_BM :=	Y1 - Y2			ctg_BM = -0.762616			a := tg_BM		c := ctg_BM
	X2 - X1
Определяем координаты точки М
S1 := a × (Y1 - YB) + ( X1 - XB)					S1 = -2806.316
S2 := a × (Y2 - YB) + ( X2 - XB)					S2 = -2806.316
XM := XB +		S1		XM = 6563.941			XM := XB +	S2		XM = 6563.941
		1 + a2						1 + a2
YM := YB+ a × ( XM - XB) YM = 11848.820
Контроль
S3 := c × ( X1 - XB) + (Y1 - YB)						S3 = 2140.142
S4 := c × ( X2 - XB) + (Y2 - YB)						S4 = 2140.142
XM := XB + c × (YM - YB)					XM = 6563.941
YM := YB+ (YM - YB)					YM = 11848.820
Заключительный контроль
b_MD := dir_ug( XM,YM, XD, YD)						b_MD = 0.328				ugol(b_MD) = ( 18 48 4)
b_MC := dir_ug( XM,YM, XC, YC)						b_MC = 5.993
										ugol(b_MC) = ( 343 21 27)
b3_k := b_MD - b_MC + 2 × p
bk_3 := ugol(b3_k)				bk_3 = (35		26 36)

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68

Пример 3

Тестовый пример 3 для

апробации программного продукта Задача: вычисление

прямоугольных координат пунктов геодезической сети, определяемых обра

засечкой (Ю.Рандярв)

b1 := ( 86 17 10)	b2 := ( 49 55 57)	b3 := ( 41 52 16)
XA := 6871.271	YA := 10373.409
XB := 9256.180	YB:= 11023.821
XC := 8979.654	YC:= 12678.256
XD := 7498.451	YD:= 12262.399

éæ

b1

, 1

b1

, 2

ö

p

ù

b1 := cotêçb1

, 0

+

0

+

0

÷

×

ú

b1 = 0.064911

60

3600

ëè

0

ø

180û

éæ

b2

, 1

b2

, 2

ö

p

ù

b2 := cotêçb2

, 0

+

0

+

0

÷

×

ú

b2 = 0.841109

60

3600

ëè

0

ø

180û

X1 := XA + (YA - YB) × b1					Y1:= YA + (XB - XA) × b1
X1 = 6829.052					Y1 = 10528.215
X2 := XC + (YB - YC) × b2					Y2:= YC + ( XC - XB) × b2
X2 = 7588.094					Y2 = 12445.667
Контроль правильности вычислений
XB_XA :=				[b1 × (Y1 - YB) - ( X1 - XB)]					XB_XA = 2384.909	XB - XA = 2384.909
				1 + b12
XC_XB:=			[b2 × (Y2 - YB) + ( X2 - XB)]						XC_XB= -276.526	XC - XB = -276.526
				1 + b22
YA_YB:=		[b1 × ( X1 - XB) + (Y1 - YB)]							YA_YB= -650.412	YA - YB= -650.412
				1 + b12
YB_YC:=	[b2 × ( X2 - XB) - (Y2 - YB)]								YB_YC= -1654.435	YB- YC = -1654.435
2

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69

Вычисляем дирекционный угол ВМ
tg_BM := X2 - X1				tg_BM = -0.395859
Y1 - Y2
ctg_BM :=	Y1 - Y2			ctg_BM = -2.526151			a := tg_BM		c := ctg_BM
	X2 - X1
Определяем координаты точки М
S1 := a × (Y1 - YB) + ( X1 - XB)					S1 = -2230.937
S2 := a × (Y2 - YB) + ( X2 - XB)					S2 = -2230.937
XM := XB +		S1		XM = 7327.479			XM := XB +	S2		XM = 7327.479
		1 + a2						1 + a2
YM := YB+ a × ( XM - XB) YM = 11787.315
Контроль
S3 := c × ( X1 - XB) + (Y1 - YB)						S3 = 5635.686
S4 := c × ( X2 - XB) + (Y2 - YB)						S4 = 5635.686
XM := XB + c × (YM - YB)					XM = 7327.479
YM := YB+ (YM - YB)					YM = 11787.315
Заключительный контроль
b_MD := dir_ug( XM,YM, XD, YD)						b_MD = 1.225				ugol(b_MD) = ( 70 12 26)
b_MC := dir_ug( XM,YM, XC,YC)						b_MC = 0.495
										ugol(b_MC) = (28 20 9)
b3_k := b_MD - b_MC
bk_3 := ugol(b3_k)				bk_3 = ( 41		52 17)

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