谁知道卡西欧4850坐标正算的程序?

2024-11-30 11:43:34
推荐回答(3个)
回答1:

CASIO4800、4850坐标计算程序[MC2007.12.30](全线贯通、全线参数程序化一次性输入,参数采用扩充变量存储,输入不需要修改程序内容,精度可控)
可以自信的说这是网上目前最好的CASIO坐标计算程序
2007.12.30重新修改上传,加入了例题;将变量输入做了一些修改,更人性化,显示NEXT,如输入下一代则继续,否则请退出;对几处上传小笔误进行了修改。
程序特点:
真正的全线贯通坐标计算,在曲线元要素输入时仅需要输入第一段全部曲线元要素,后面曲线元要素除起点半径、终点半径、曲线长、转向需输入外其他要素均从前一曲线按辛普森8等分计算得出;辛普森公式任意等分,满足所有精度要求;全线曲线元数据一次性程序化输入,扩充变量数据库,无需修改程序内容;傻瓜操作,适用初级用户;

回答2:

. 主程序 ZFS
"1. KL => XY":"2.XY => KL ":N:I"X0":S"Y0":O"K0":G"F0":H"KN":P"R0":R"RN":Q”Q(-Z +Y)” :
D=(P-R)÷(2(H-O)PR): N=1=>Goto 1:≠>Go to 2Δ←┘
Lbl 1:{KL}:KL:J=Abs(K-O):Prog "SUB3":”X=”:X=X◢”Y=”:Y=Y◢
”FWJ=”:F=F-90◢
Goto 1←┘
Lbl 2:{ XY }:XY:Z[2]=X:Z[3] =Y: Prog "SUB4":
”K=”:K=O+J:”L=”:L=L◢
Goto 2

2. 正算子程序(SUB3)
Defm 4:
A=0.1184634425:B=0.2393143352:Z[4]=0.2844444444:C=0.0469100770:E=0.2307653449:Z[1]=0.5:
X=I+J(Acos(G+QCJ(1÷P+CJD)×180÷π)+Bcos(G+QEJ(1÷P+EJD)×180÷π)+Z[4]cos(G+QZ[1]J(1÷P+Z[1]JD)×180÷π)+Bcos(G+Q(1-E)J(1÷P+(1-E)JD)×180÷π)+Acos(G+Q (1-C)J(1÷P+(1-C)JD) ×180÷π)):
Y=S+J(Asin(G+QCJ(1÷P+CJD)×180÷π)+Bsin(G+QEJ(1÷P+EJD)×180÷π)+Z[4]sin(G+QZ[1]J(1÷P+Z[1]JD)×180÷π)+Bsin(G+Q(1-E)J(1÷P+(1-E)JD)×180÷π)+Asin(G+Q (1-C)J(1÷P+(1-C)JD) ×180÷π)):
F=G+QJ(1÷P+JD) ×180÷π+90:X=X+LcosF:Y=Y+LsinF

回答3:

一、程序:MC
W“1.JS 2.SZ”:W=1=>Z[2]=0:Goto 1 ΔW=2=> O“KOU LING”:O≠123456=>O=0: “OUT” ◢Goto 4ΔO=0: Z[1]=0:Goto 0←┘
Lbi 0←┘
”N0.”:Z[1]+1 ◢←┘
Z[1]=0=>{ABCREFGU}:A“X0”:B“Y0”:C“F0”:R“R0”:E“RN”:F“D0”:G“LS”:U“G” : Z[Z[1]×8+3]=A:Z[Z[1]×8+4]=B:Z[Z[1]×8+5]=C:Z[Z[1]×8+6]= R-1:Z[Z[1]×8+7]= E-1:Z[Z[1]×8+8]=F: Z[Z[1]×8+9]=F+G: Z[Z[1]×8+10]=U: “NEXT”◢ Isz Z[1]: Goto 0ΔZ[1]=1=>D=Z[9]:Z=0:Z[2]=0:GOTO 2ΔD=Z[(Z[1]-1)×8+9]:Z=0:Z[2]=Z[1]-1:GOTO 2←┘
Lbi A: Z[Z[1]×8+3]=X:Z[Z[1]×8+4]=Y:Z[Z[1]×8+5]=J: Z[Z[1]×8+8]=D: {REGU}:R“Ro”:E“RN”: G“LS”:U“G”: :Z[Z[1]×8+6]=R-1 Z[Z[1]×8+7]=E-1: Z[Z[1]×8+9]=D+G: Z[Z[1]×8+10]=U: “NEXT”◢ Isz Z[1]: Goto 0←┘
Lbi 1←┘
{DZ}:D:Z:Z[2]=0:Goto 2←┘
Lbi 2←┘
Z[2]>Z[1]-1=>GoTo 4ΔD≤Z[Z[2]×8+9]=>A=Z[Z[2]×8+3]:B=Z[Z[2]×8+4]: C =Z[Z[2]×8+5]:R=Z[Z[2]×8+6]: E=Z[Z[2]×8+7]: F=Z[Z[2]×8+8]: G=Z[Z[2]×8+9]: U=Z[Z[2]×8+10]: Goto3ΔIsz Z[2]:Goto 2
Lbi 3←┘
W=2 =>N=8:≠>N=5ΔP=U(E-R)÷Abs(G-F)÷N:Q=Abs(D-F):S=90Q÷π: J=C+(NPQ+2UR)NS:L=1←┘
X=A+Q÷6×(Cos C+Cos J +4∑(Cos (C+((L+0.5)PQ+2UR)×(L+.5)S),L,0,(N-1))+2∑(Cos (C+((LPQ+2UR)LS,L,1,(N-1)))+ZCos(J+90)←┘
Y=B+Q÷6×(Sin C+Sin J +4∑(Sin (C+((L+0.5)PQ+2UR)×(L+.5)S),L,0,(N-1))+2∑(Sin (C+((LPQ+2UR)LS,L,1,(N-1)))+Z Sin(J+90):W=2=>GOTO AΔZ=0=>“X”:X:Pause 0: “Y” :Y◢ Goto 1Δ
Z<0=>“XL”:X:Pause 0: “YL”:Y◢ Goto 1Δ fx4850 ①
Z>0=>“XR”:X:Pause 0: “YR”:Y ◢ Goto 1 ←┘Z=0=> X “X” ◢Y “Y”◢ Goto 1Δ
Z<0=> X “XL” ◢Y “YL”◢ Goto 1Δ fx4800 ②
Z>0=> X “XR” ◢Y “YR”◢ Goto 1 ←┘
Lbi 4←┘

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