GPS單點(diǎn)定位實(shí)驗(yàn)報(bào)告

1、GPS原理與應(yīng)用實(shí)驗(yàn)題目 GPS單點(diǎn)定位專 業(yè)：測繪工程班 級： 12-01學(xué)號：2012212600姓 名：王威指導(dǎo)教師：陶庭葉時間：2014.11目錄一、實(shí)驗(yàn)?zāi)康?3二、實(shí)驗(yàn)原理3三、實(shí)驗(yàn)內(nèi)容3四、實(shí)驗(yàn)效果圖 9五、實(shí)驗(yàn)總結(jié) 9一. 實(shí)驗(yàn)?zāi)康?. 深入了解單點(diǎn)定位的計(jì)算過程；2. 加強(qiáng)單點(diǎn)定位基本公式和誤差方程式，法線方程式的記憶；3. 通過上機(jī)調(diào)試程序加強(qiáng)動手能力的培養(yǎng)。二. 實(shí)驗(yàn)原理一個接收機(jī)接受三個火三個以上衛(wèi)星信號，得出衛(wèi)星坐標(biāo)和偽距, 利用間接平差計(jì)算接收機(jī)的坐標(biāo)。三. 實(shí)驗(yàn)內(nèi)容1. 程序流程圖2、實(shí)驗(yàn)數(shù)據(jù)某個歷元衛(wèi)星坐標(biāo)（單位：“PRNXYZ-652183. 561823536

2、896. 6765-10392867, 6529VF Vj-12412058, 257822624932. 2093-5728436.9498G16,-3373128. 604822004627. 781014303822.4050G曲101249&2. 782011227597. 805321807939. 5847G31,-17262027. 67719118996.109118295569. 1090偽距觀迴數(shù)據(jù)PRN偽距（）鐘差（s）G 323298409, 6000. 00059981G &22041347,5800, 00051935G1620498932. 300

3、-CL 0001026G2323344899. 9600.00035684G3121424389. 460-1. 4372e-53、實(shí)驗(yàn)程序代碼Private Sub Comma nd1_Click()Commo nDialog1.Filter = "TXT files|*.txt|"Common Dialog1.FilterI ndex = 1Common Dialog1.ShowOpe nOpe n Me.Com mon Dialog1.FileName For In put As #1Do While Not EOF(1)Line In put #1, Texttex

4、tbuff = textbuff + Text + vbCrLfLoopClose #1kk = MSFIexGrid1.Rows - 1Dim aReDim a(kk - 1)a = Split(textbuff, vbCrLf)For j = 1 To kkFor i = 1 To 5 MSFlexGrid1.TextMatrix(j, i) = a(j - 1 + 5 * (i - 1)Next iNext jFor k = 1 To kkMSFlexGrid1.TextMatrix(k, 0) = "第" & k & " 個點(diǎn)"N

5、ext kMSFlexGrid1.TextMatrix(0, 1) = "X"MSFlexGrid1.TextMatrix(0, 2) = "Y"MSFlexGrid1.TextMatrix(0, 3) = "Z"MSFlexGrid1.TextMatrix(0, 4) = "偽距 "MSFlexGrid1.TextMatrix(0, 5) = "鐘差 "End SubPrivate Sub Command2_Click()kk = MSFlexGrid1.Rows - 1X0 = 0: Y0

6、= 0: Z0 = 0c = 299792458Dim a()ReDim a(kk - 1, 3)Dim ll()ReDim ll(kk - 1, 0)For ii = 1 To 100For i = 1 To kkl = (MSFlexGrid1.TextMatrix(i, 1) - X0) / Sqr(MSFlexGrid1.TextMatrix(i,1)-X0) A 2+ (MSFIexGrid1.TextMatrix(i, 2)-Y0) A 2(MSFIexGrid1.TextMatrix(i, 3) - ZO) a 2)m = (MSFIexGrid1.TextMatrix(i, 2

7、) - Y0) / Sqr(MSFIexGrid1.TextMatrix(i, 1)- XO) A 2+ (MSFIexGrid1.TextMatrix(i, 2)- YO) A 2(MSFIexGrid1.TextMatrix(i, 3) - ZO) A 2)n = (MSFIexGrid1.TextMatrix(i, 3) - ZO) / Sqr(MSFIexGrid1.TextMatrix(i, 1)- XO) A 2+ (MSFIexGrid1.TextMatrix(i, 2)- YO) A 2(MSFIexGrid1.TextMatrix(i, 3) - ZO) A 2)a(i -

8、1, O) = Ia(i - 1, 1) = ma(i - 1, 2) = na(i - 1, 3) = -1Ik = MSFIexGrid1.TextMatrix(i, 4) - Sqr(MSFIexGrid1.TextMatrix(i, 1) XO) A 2+ (MSFIexGrid1.TextMatrix(i,2)- YO) A 2(MSFIexGrid1.TextMatrix(i, 3) - ZO) A 2) + c * MSFIexGrid1.TextMatrix(i, 5)II(i - 1, O) = IkNext igzs = xc(qiuni(xc(zz(a), a), xc(

9、zz(a), II)X0 = X0 - gzs(0, 0)Y0 = Y0 - gzs(1, 0)Z0 = Z0 - gzs(2, 0)j = j + 1Next iiText2.Text = "X=" & X0 & vbCrLf & vbCrLf & "Y=" & Y0 & vbCrLf & vbCrLf& "Z=" & Z0V = jian(ll, xc(a, gzs)zjl = xc(zz(V), V)(T 0 = Sqr(zjl(O, 0) / (kk -

10、3)Qx = qiuni(xc(zz(a), a)Text3.Text = " c X=" & c 0 * Sqr(Qx(O, 0) & vbCrLf & vbCrLf & "c Y=" &(T 0 * Sqr(Qx(1, 1) & vbCrLf & vbCrLf & "c Z=" & c 0 * Sqr(Qx(2, 2)End SubPrivate Sub Form_Load()MSFlexGrid1.ColWidth(1) = 1300MSFlexGrid1.

11、ColWidth(2) = 1300MSFlexGrid1.ColWidth(3) = 1300MSFlexGrid1.ColWidth(4) = 1300Text2.Text = ""Text3.Text = ""End Sub' 矩陣相減Public Function jian(m, n)Dim i, j As IntegerIf UBound(m, 1) <> UBound(n, 1) Or UBound(m, 2) <> UBound(n, 2) ThenMsgBox (" 請確認(rèn)輸入數(shù)組是否可以相減！

12、")ElseDim c()ReDim c(UBound(m, 1), UBound(n, 2)For i = 0 To UBound(c, 1)For j = 0 To UBound(c, 2)c(i, j) = m(i, j) - n(i, j)Next jNext ijian = cEnd IfEnd Function' 矩陣的轉(zhuǎn)置Public Function zz(a)Dim i As Integer, j As Integer, t As Integer, b()If UBound(a, 1) = UBound(a, 2) ThenFor i = 0 To UBou

13、nd(a, 1)For j = 0 To UBound(a, 2) If i < j Then t = a(i, j) a(i, j) = a(j, i) a(j, i) = t End If Next j Next i zz = a ElseReDim b(UBound(a, 2), UBound(a, 1) For i = 0 To UBound(a, 2) For j = 0 To UBound(a, 1) b(i, j) = a(j, i)Next j Next i zz = b End If End Function ' 兩矩陣相乘 Public Function xc

14、(a, b) Dim i As Integer, j As Integer, k As Integer If UBound(a, 2) <> UBound(b, 1) Then MsgBox (" 這兩個矩陣不能夠相乘 ") Exit FunctionEnd IfReDim sd(UBound(a, 1), UBound(b, 2) For i = 0 To UBound(a, 1)For j = 0 To UBound(b, 2) For k = 0 To UBound(b, 1) sd(i, j) = sd(i, j) + a(i, k) * b(k, j)

15、 Next k Next j Next i xc = sd End FunctionPublic Function qiuni(a) Dim c, m%, n%, p#, l%, i%, j%, ab# m = UBound(a, 1) n = UBound(a, 2) If m <> n ThenMsgBox (" 該矩陣不可逆！ ") Exit FunctionEnd IfReDim c(m, 2 * n + 1)For i = 0 To m For j = 0 To n c(i, j) = a(i, j) Next j Next i For i = 0 T

16、o m For j = m + 1 To 2 * m + 1 c(i, j) = 0 Next j Next i i = 0For j = m + 1 To 2 * m + 1 c(i, j) = 1 i = i + 1 Next jFor k = 0 To n If c(k, k) = 0 Then For i = k + 1 To n If c(i, k) <> 0 Then GoTo this End If Next i If i = n + 1 Then MsgBox (" 該矩陣不可逆 !") Exit Function End If this:For

17、 j = 0 To 2 * m + 1 p = c(k, j) c(k, j) = c(i, j) c(i, j) = p Next j End If ab = 1# / c(k, k) For j = 0 To 2 * m + 1 c(k, j) = c(k, j) * ab Next j For i = 0 To n If i <> k ThenFor j = 0 To 2 * m + 1If j <> k Thenc(i, j) = c(i, j) - c(i, k) * c(k, j) End IfNext jc(i, k) = 0End IfNext iNex

18、t kFor i = 0 To mForj = 0 To ma(i, j) = c(i, j + n + 1)a(i, j) = Round(a(i, j), 4)Next jNext i qiu ni = a End Fun cti on四. 實(shí)驗(yàn)結(jié)果圖廚亦i， =J 回 | S3 ?ITZH-2562064.17S5663Y=48C7656.004236142=3226573.2115236471=1.007426316071(7Y-2.00326992270373<7 Z=. 021616640622JS133,561"2353MK.676510392867.65299409.BOOJ0Q59981I112058. sra22624932 2093-5723136. W