JPH0478961A - Graphic processor - Google Patents

Abstract

PURPOSE:To discriminate a fact that a point on the layer exists on the layer by deriving a characteristic line for showing the surface of the layer, based on information for showing the layer existing in a solid space, information for designating a point on the layer, and information for showing the direction for looking at the layer, adding it to the part related to the point on the layer and displaying it. CONSTITUTION:A first means 1 of a processing part 3 calculates a point on the layer by a method by which a point in which a straight line for passing through a point on a screen and being parallel to the eye direction intersects with the layer, and from the layer and information for designating one point on the layer. Subsequently, a second means 2 of the processing part 3 adds a characteristic line for showing the surface of the layer to the part related to one point on the layer and displays it on a display part 5, from the point on the layer obtained as a result of a first means 1 and the eye direction. In such a way, whether the displayed point exists on the layer or not can be discriminated with eyes.

Description

Detailed Description of the Invention [Industrial Field of Application] The present invention enables the display of points on a specified surface by adding and displaying a characteristic line representing the surface of the surface. This invention relates to a graphic processing device that can identify the presence of a surface on a surface.

[Prior Art] Conventionally, points in three-dimensional space are displayed by showing the position of the point on the screen.
For example, an X mark was displayed. Note that this type of technology is disclosed in Japanese Patent Application Laid-Open No. 62-226376 and the like.

[Problems to be Solved by the Invention] The above-mentioned conventional technology has a problem in that it is not possible to identify which surface the displayed point is on. An object of the present invention is to provide a graphic processing device that can identify the presence of a point on a surface.

[Means for Solving the Problems] In order to solve the above problems, at least in a graphic processing device having an input section and a display section, information indicating a surface existing in a three-dimensional space inputted from the input section and a surface Based on the information specifying the point above and the information indicating the viewing direction of the surface in 3D space, obtain the point on the surface, find a characteristic line representing the surface of the surface, and connect the point on the surface. What is necessary is to add it to the relevant part and display it on the display section.

[Operation] Based on the information indicating the surface existing in the 3D space, the information specifying the point on the surface, and the information indicating the direction in which the surface of the 3D space is viewed,
Obtain a point on the surface, find a characteristic line representing the surface of the surface,
By adding it to a portion related to a point on the surface and displaying it on the display section, it is possible to identify that a point on the surface exists on the surface.

[Example] An example of the present invention will be described in detail below with reference to the drawings.6
FIG. 1 shows an example of the configuration and data flow of a graphic processing device to which the present invention relates.

This embodiment is composed of an input section 4, a display section 5, and a processing section 3. The processing section 3 includes a first means 1 for obtaining points on a surface from information specifying a surface and a point on the surface, It is configured to include a second means 2 for adding a characteristic line representing the surface of the surface from the upper point and the viewing direction as necessary to the point on the surface and displaying it using the viewing direction. Note that the first means is used to obtain points on the surface from the information specifying the surface and points on the surface.The second means is to display points on the surface by adding characteristic lines representing the surface of the surface. It is said that The input unit 4 includes a keyboard, a mouse, a light pen, a display, a tablet, etc., and the display unit 5 includes a display and the like.

FIG. 2 shows when the graphic processing device according to the present invention is used,
This is an example of the shape displayed on the display screen.

First, let us assume that one surface existing in three-dimensional space is given. An example is shown in FIG.

Next, assume that one point on the screen is given as information specifying one point on the surface, for example in the format shown in FIG. 4.

Next, the direction in which the surface shown in Fig. 3 is viewed (line of sight direction) is
FIG. 5 shows an example of displaying on the screen using . It is common for graphic processing devices that store and display figures in a three-dimensional space to display a surface existing in a three-dimensional space by giving a line of sight direction.

Here, the first means 1 of the processing unit 3 uses the screen of the example of FIG. A point on the surface is calculated by using a method in which a point on the surface is defined as a point on the surface where a straight line passing through the above point and parallel to the viewing direction intersects with the surface shown in the example of FIG. Its form is shown in FIG. Although the point on the screen and the line of sight are used as information to obtain the point on the surface, this example is not the only method for determining the point on the surface. Then, this may be set as a point on the surface, or alternatively, a distorted line may be drawn onto the surface from the coordinate values based on the numerical values, and the foot of the obtained perpendicular line may be set as a point on the surface.

Next, the second means 2 of the processing unit 3 represents the surface of the surface from the surface (for example, the example shown in FIG. 3), the points on the surface obtained as a result of the first means 1, and the viewing direction. A characteristic line is displayed by adding it to a portion of the surface corresponding to one point on the surface, as shown by the broken line in FIG. 7, for example. In the case of FIG. 7, the direction is added along the shape of the surface from one point on the surface toward the side constituting the surface, with the direction parallel to the side. It can be understood from FIG. 7 that one point on the surface appears to be on a surface composed of quadrilaterals connected by broken lines.

Another example is shown in FIG. FIG. 8 shows a case where one point on a surface existing on a spherical surface is expressed. Both are expressed using two lines on the surface passing through one point on the surface. The above characteristic line has this meaning.

Next, the first means 1 will be explained. The first means 1 of the processing unit 3 calculates one point on the surface from a surface (for example, the example in FIG. 3) and information specifying one point on the surface (for example, the example in FIG. 4). . This will be explained below using an example. Also, the second
The flowchart in Figure 4 will be used for explanation. Here, the surface (
(see FIG. 6), the following values are given. In the flowchart of FIG. 24, this is shown in step 21. The origin on the plane is P. , the first axis indicating the reference direction is P. As the direction from P2 to u(lul=1
), the second axis perpendicular to the first axis, P in FIG. From P,
Let V (lVl=1) be the direction toward , and let each of these elements be expressed as follows.

1st axis......u = (Ux, Uy, Uz)
・・・・・・・・・(1) Second axis・・・・・・・・・
・v = (Vx, Vy, Vz) ・・・・ ・・
・(2) Origin・・・・・・・・・P, =(Xo,
(Yo, Zo) ・・・・・・・・・・・・(3) Also, the following value shall be given for the viewing direction.

Line of sight direction...k = (Kx, Ky, Kz)
・・・・・・・・・・・・(4) Also, the point on the screen in the example of Fig. 4 (position of the line of sight on the screen) is as follows:
The following values shall be given.

Point on the screen...p, = (x-work, Yl, Z□)...
(5) Under the above conditions, a straight line parallel to the line of sight direction is P shown in the example of FIG. , P2, P3
When calculating by a method in which a point that intersects with a surface determined by three points is regarded as one point on the surface, the procedure is as follows.

First, from equations (1) and (2), the third axis W, which includes the first, second, and third axes and constitutes a right-handed system, is determined using the following equation. In the flowchart of FIG. 24, reference is made to step 22.

3rd axis...w=uxv...
(6) Expressed in component values, it has the following form.

If w = (Wx, Wy, Wz) ---------(7), then Wx = UyVz −UzVy ---(8)Wy
= UzVx −UxVz ・・・・・・−−(9)
Wz=UxVy−UyVx−−−−−=−(10). Now, by using equations (1), (2), (3), and (6), the U-v-w coordinate system has been created.

Here, the coordinate values of the line-of-sight position P1 on the screen in equation (5) are converted to values in the u-v-w coordinate system. In the flowchart of FIG. 24, step 23 is referred to.

The value of P□ in the u-v-w coordinate system is (P□u, Plv, P
ow), it becomes the following form.

pHJ= (Lx-Xo)Ux+ (Yx-Yo)L
Iy+ (zi-Zo)Uz・・・・・・・・・・・・(
11) PXv= (Xl-Xo)Vx+ (Y□-Yo)Vy
+ (Z□-Z,)Vz... (12) P1w= (X, -X.)ldx + (Y, -Y,
)Wy + (Z□−Zo)suZ・・・・・・・・・
-(13) Next, the vector k indicating the line-of-sight direction is converted to a value in the u-v-w coordinate system. In the flowchart in Figure 24, step 2
See 4. If the values of the vector in the U-v-w coordinate system are (Ku, Kv, Kw), the following form is obtained.

Ku=Kx・Ux+Ky−Uy+Kz−Uz
--(14) Kv=Kx-Vx+Ky-Vy+Kz-V
z ---... (15) Kw=Kx-Wx+Ky-
Wy+Kz-Wz ・=----=・(16) Here, when the point Pi on the straight line passing through the line-of-sight position P1 on the screen and facing the vector indicating the line-of-sight direction is on the surface, Pi Find the value of. The value of Pi is the value of one point on the surface to be determined in this example.

In the flowchart of FIG. 24, step 25 is referred to. In step 25, only the formula to be determined is shown.

The process to obtain this formula will be described below. A point Pi on a straight line passing through the line-of-sight position P on the screen and facing the vector indicating the line-of-sight direction can be expressed in the following form.

Pj=a-+Pe ・・・・・・・・・・・・(17
) If this is expressed in terms of each coordinate value, and if the values of Pi in the u-v-w coordinate system are (Piu, Piv, Pitll), then the following form is obtained.

Piu=a−Ku+P1u −−・・−・(18)P
iv=a−Kv+P,v・・−=・−・(19)P
iIll=a-nisu+P engineering w・・・・・・・・・・・・
(20) Here, the point where the point Pi on the straight line intersects with the u-v plane is Pi = 0.
...This is the point in (21). If formula (20) is used under this condition, a will have the following value.

a = -P1 enemy/Kw ・・・・・・・・・・・・
・(22) From this value of a, (P
iu, Piv) can be obtained in the following form using equations (18) and (19).

Piu=(-P1ty/Kw)4u+P, u --・
・-・(23) Piv=(-P1w/Kw)4v+P,
v −... (24) Here the obtained (Piu, Pi
v) is a value in the u-v-w coordinate system. This is x-y
When converted to a value in the −z coordinate system, it becomes as follows. Second
In the flowchart of FIG. 4, reference is made to step 26. In this case, (Piu, Piv
, O) can be returned to the value in the x-y=2 coordinate system, so
If P] is (Xi, Yi, Zi), then
X1=Piu-Ux+Piv-Vx+Xo--(25)
Yi=Piu-Uy+Piv-Vy+Yo--(26)
Zi=Piu(Iz+Piv-Vz+Z.---(27
) is obtained in the form.

When using one point on a surface as a value projected onto a two-dimensional plane, such as when using it as a value on a display screen, the above results (Piu, Piv) are expressed as 3 as representing the position in three-dimensional space. When used as dimension values, the above results (Xi, Yi, Zi) may be used.

Next, the second means 2 will be explained. The second means 2 of the processing section 3 processes a surface (for example, the example shown in FIG. 3) and a point on the surface obtained as a result of the first means.

From the line of sight direction, a characteristic line representing the surface of the surface is added to a portion of the surface related to one point on the surface and displayed.

This will be explained below using an example.

First, the case of FIG. 7 is an example where the surface is square or rectangular in three-dimensional space. The flowchart in FIG. 25 will also be used for explanation. In this case, as shown in FIG. 7, an image line segment is added as a dashed line from one point on the surface toward the side constituting the surface, the direction of which is parallel to the side adjacent to the side toward which it is directed. In this case, the information given includes the following (see FIG. 10). In the flowchart of FIG. 25, step 31 is referred to.

As one point on the surface, Pi =: (Xi, Yi, Zi) - (30
) As the coordinate values of the vertices of the rectangle, P2=(X,,Y2.Z2) -=-(:n)p3=
(X,,Y3-Z3) -・-・=(32)P,=(
X,,Y,,Z4) ==...--(33)p,=
cx9. vs, zs) ・−・・−・(34) Also,
What is obtained as a result is, for example, as shown in FIG. 11, the end points P6, P7, pIl, p,
There is. Let them be P, = (Xs, YG, ZG) ・
==-・-(35)P7=(X7, Y7, Z
7) ・・・・・・・・・(36)P8=(X,,Y
,,Z8) ・・・・・・・・・(37)P9=(X,
, Yi, Z9) =--old... (38).

First, the unit vector W□ moves from P2 to P5.
, P2 to P, a unit vector w2 is created.

In the flowchart of FIG. 25, reference is made to step 32. The component values of W and w2 are expressed as wx = (Wxx tw
yt rwzl) -...= (39) W2 =
(Vx, tWyz twz2) ”'”'”'
(40), then wkl==X, x2-...old-(41)wk2=
Y5-Y2・・・・・・(42)vk3=Z,
−22−−−(43) wk4= v' (wk12+w
k2"+tyk32)--" (44) Wx□=wk
l/1gk4 -----------(4s)Wy,
= wk2/ wk4 --...(46)W
z, = wk3/wk4 − (47) wk5 = x,
-X2・・・・・・(48) tllk6 = Y,
-Y2......(49)vk7
= Z, −Z2・= ・・・・・・(50) 8
= J (tllk52+wk62+suni72)・
... (51) Wx, = wk5 / wk8
--- (52) SuYz = 6 for enemy / 8 for υ
・・・・・・・・・(53)suZ2= tllk7 /
tik8... can be obtained by the procedure (54).

Next, P2=(X2.Y2.Z2) is the starting point, Pi=(X
i+Yi.

A direction vector W3 having Zi) as the end point is determined. 25th
In the illustrated flowchart, reference is made to step 33.

W3 = (wx, tllya, Wx3) ・=
−・−(55), then enemy x3=Xi−X2 ・・・・・・・・・(5
6) Wya = Yi −Y2 −・−−−−−
--(57)Wz,=zi-Z2・・・・・・・・・(
58). Here, W□ of the direction vector with P2 as the starting point and Pi as the ending point, component values d1, d in the W2 direction
Find 2. In the flowchart of FIG. 25, reference is made to step 34. , their dl and d2 can be obtained in the following form.

d, =Wx3・Wx1+Wy,・Wy,+li'z,
・Wzl−・−・(59)d2=Wx3・Wx2+1
jy3・Wy2+Ijz3・Wz2−・−・・(60)
Using d□ and d2 obtained here, P, , P7, P3
,P, can be obtained in the following form. In the flowchart of FIG. 25, reference is made to step 35.

P, = d1・W1+P2 ・・・・・・・・・(61
)P7=d2・W2+P2 ・・・・・・・・・(
62) P, = d□・W engineering + P3 ・・・・・・・・・
(63) P9=d2・W2+PS ・・・・・・・・・
・(64) When this is expanded into each coordinate value, it becomes the following form.

X, =d1・Wx1+X, -(65)Y,
= d engineering・wy1+y2 ・・・・・・・・・(6
6) Z, =d16Wz1+Z2...-467)x7
=d2−IIlx2+x, −・−−168)
Y, =d2・νy2+Y2・・・・・・・・・
(69) Z7=d2・Wz2+z2・・・・・・・・・
(70)
...(71)Y, = d1・-y engineering+Y3
... (72) Z, =d1・Wz, +Z
, −−・=−(73)Xg=d2・Wx2+
X, ---(74)Y, =d2・Wy2
+Y, ...... (75)zg=d2
・WZ2+Z,...(76) Then, p...
By displaying the line segment connecting p8 as a broken line and the line segment connecting P7 and P9 as a broken line, the desired result can be obtained. In the flowchart of FIG. 25, reference is made to step 36.

Next, the case of FIG. 9 is an example of a general quadrilateral whose opposing sides are not parallel. In addition, the flowchart in FIG. 26 will be used for explanation. In this case, the information given includes the following:

In the flowchart of FIG. 26, step 41 is referred to. As one point on the surface, pi==(xi, yi, zi) ・−・−・−・−・
(100) As the coordinate values of the vertices of the quadrilateral, PZ” (X2.Y2-22) ・・・・・・・・・(
101) P3=(X3.Y3.Z,) ・・・・・・
...(102)P4=(X4.Y4.Z4) ...
......(103)P5=(X9.Y9.Z,)
--- (104) What is obtained as a result is, for example, as shown in FIG. 9, there are two end points P0, P7, and P of the added broken line. P6=(X6.YG,Z,) ・・・・・・・・・(1
05) P7= (X7.Y, , Z7)...
...(106)P8=(X,l,Y8.Z,)
-・・・・・・・・・(107)P9”(X9.Y9.Z9
) """"' (108).

First, in order to represent the corresponding positions when equally divided on opposing sides P2P, P, and P4, the length when each is divided into n equal parts is determined. In the flowchart of FIG. 26, reference is made to step 42. Let d3 be the length when sides p, p are divided into n equal parts, and d4 be the length when side P5P4 is divided into n equal parts, then d3 and d4 can be obtained by the following equations. Furthermore, n
is a variable that disappears when the resulting formula is obtained, but when constructing a surface, there is a method of dividing two lines by the same equal number and connecting the corresponding lines. Therefore, I used it.

d, = IP3-P2I/n (109)
d4==l P4- P, l/n --(11
0) Next, the unit vector w going from P2 to P
,, create a unit vector W from p to P4. Please refer to FIG. Note that in the flowchart of FIG. 26, step 43 is referred to.

The component values of W3 and W4 are Wz ”= (WX3 Jy31111Z3) −'
”-...(111)W4=(WX41Wy41WZ
4) -...Mountain- (112), then vkl=X, X2---(113)tik2=Y,
-Y2 ・・・・・・・・・(114) wk3
:Z3 z2 +++ ・++ +++ (
115) wk4=v' (wk12+tgk22+w
k32)・=”・river(116)vx3=tykl/
wk4 --- (117)Wy, ==wk2
/ wk4 ・= ・−-(118) Wz3=
wk3/ wk4 --- (119) wk5=X
, -X, ...old...(120)wk6
= Y4- YS---(121) wk7 = Z,
-Z, =・−・−−−−(122) ri 8
= J (wk52+suni62+ω72)・
(123) Wx, = tyk5 / tyk8
・−・・・・(] 24)Wy4= wk6/ w
k8 --- (125) lJz, = wk7/
It is obtained by the procedure wk8 .---(126).

Using d3 and W obtained above, a point Pw on side P2P3 with P2 as a reference can be obtained in the following form.

In the flowchart of FIG. 29, reference is made to step 44.

Pty3=b-d3W3+P2 ・・・・・・・・・(
], 27) Also, using d4 and W4 obtained above, P.

A point Pw4 on the side P5P4 based on P5P4 is obtained in the following form.

Pw4=b-d4W4+P, --(128) Here, line segment Pv3Pw4 is one point P on the given surface
If the variable is set so that it passes through i, one desired line segment can be obtained. In other words, line segment Pw.

What is necessary is to determine the conditions for Pj to be on Pw4. To do this, a coordinate system with Pi3 as the origin and a reference axis in the direction from Pw3 to Pw may be introduced, and the determination may be made based on the distance of Pi from the reference axis.

First, the origin is Ptv3=b-d, W, +P2-・-(+29), then,
The reference axis is V/, =P1.14-Pen, ... (1,3
0)=b(d4W,-c13W,)+PS-P.

To determine the degree of separation, an axis perpendicular to the reference axis is required, and if this axis is W, Wr can be obtained by the following procedure.

W b ” W s X W 3 ・・・・・・(
131) Here, W5 and W6 are WS = (Wx61sys+WZ5) """"'
(132) We = (Wx, ,WyGtWZ6)
---(133), first, each component value of W is obtained in the following form.

Wxs = b (d 4Wx4-d 3x3)+
CX, -X2)・-・-= (134)Cy,=b(d
4sy4d a'l')'z)+ (YS Y2)・
...(135) SuZs = b (d 4Wz4
−d 3Wz, )+ (Z, −Z2)−−(136)
Next, each component value of W6 is obtained in the following form.

WxG:Wy, Wz, -WzsWy, -・-=・=
・(137)Wy,=WzJx3−Wx,Wz3---
(138) Wz, =Wx5Wy, -Wy, Wx3---
(139) The reason W3 is used here is to find a vector orthogonal to W5, and any vector may be used as long as it is not parallel to W and has a known value. Once WG is obtained, b can be found using the following equation.

(P i −P w3)・W, = O−−・・(14
0) Here, the values of each component of W6 are as follows.

IJx6=Wy, Wx3-1i1z, Wy.

” (b (d JY* d 3Wyl) + (YS
Y2)) WZ3(b (d 4'jZ4 d 3
WZa)+(zs Zz))Wya=b”d4(Wy
4Wz3WZJy3)+(Ys Y2)ll'Z3(
zs L)Wy3 ・・・・・・(1
41) Wy, = Wz, bx, -111X, W
z.

” (b (d 4WZ4 d 3WZ3) + C
10-Z2))WXj(b (d4wx4a JX+)
+(XS xz)) wzi= b-d 4(Ilz4
Wx3-INx4wz3)+(Zs-Z2)Wx.

(XS X2) WZ3 ・・・・・・・・・
...(142)Ijz6=Wx5Wy3-Wy,'
Jx3” (b (d 41i1x4d 3LJXa)
+ (Xs L))Il'ya(b (d Jy*
d xWyi)+ (Ys Y, +))WX+=
b-d4 (Wx, Wy, '74'X3) ten (Xs
X2)Wy3(Ys Y2)Wx3---(14
3) Now, if we create the scalar product (Pi Pw, )·W, we get the following.

(Pi-Pw,)" (Xi, Yi, Zi) "...-
(144) Pw3=b-d, W3+P2 ---...
(145) First, (Pi −Pw,)=(Xw3.Yw,,Zw,)−・
-.(146), each component value becomes linear.

Xw,=Xi-b-d3Wx3-X2=(146)Yw
3=Yi-b-d3Illy3-Y2・・・・・・・・・
・(147)Zti,=Zi-b-d,1ilz3-Z
2-(148) Here, (P i −P W3)·w.

= Xw3Wx, +Yw, Wy6+Zw3Wz,
--(149), so if we show each term, Xw3sx6 = (Xi b-d3wx3
'y4wZ:+WZ4Wya)+(Y,-Y2)W
z3=(Z5-Z2)Wy,) ------(150
)=b2(dad4(WxJz,Idy4WX3Wy3
IIZ4)) + b (-d 3Wx, Wz, (y,
-YZ)+ d aWx3Wy, (z5-22)+
a 4(Wz4WL+-1'yJzJ(Xi-L))+
(Xi-XZ) ((YS-Yz)WX3- (zs
-22) WX3) w3Wys = (Yx b”diWyi YZ) (b-d4(W
zJxl Wx4Wz3) ten (Zs-Zz) Wx3(
Xs -XZ)titz3) """"' (151)
= b2(d3do(Wx, WyaWz4−Wy3WZ
rWXj) 10b (-d 31AX3Vy3 (25-
z2) + d 3'JY3'JZ3 (XS -XZ)
+ d, (Yi - Yz) (Wx, Wx4 - Wz3
Wx4))+ (Yi-Yz)((Zs-Zz)Wx3
- (XS-XZ)Wz3)Zw, WzG = (Zi-b -d 31i1zi-zz) (b -
d, (Iilx4Wy, -1i1y, WX,)+(
xs-xz)WX3-(vs-yz)wx3) ・
−−−・・・−・(152)=b”(d, d4(Wz,
WyJx4 W;hWXiWyJ)+b(a3wy3w
z3(x5 XZ)+d3Wx, Wz3(Y, Y
Z) + d4 (ZjZz) (WX4Wy3Wy+Wx
3))+ (zi zz) ((xs L)Wya
(YS yz)'1x3) Here, the scalar product (X
ll1. Wx, + Yw, (ily6+ Ztg
3Wz, ), Xw]Wx, +Yw, Wy6 +Zw, ljz.

= b2(d3d, (WX, WZ, Wx4-WX, W
y, 1JZ4)) + b”(d, d4(Wx, Wy3
1dz4-Wy, Suz, Wx, ))+b”(d3d4(
-Zahal & X4 WZ3WX, WYj) 10b (-
d 3Wx, wx3(YS-YZ)+ d, WxJy
3(Z, -Z2)+d4(WX31'yq
I')'z 'Z4) (XI XZ)) 10b
(-d 3WX3Wyi (zs-Z2)+ d il
l'yJZi (xs -XZ)+d4(Yi Yz
)(Wx3Wz4-Wz3Wxj)+b (-d 3
Wy, WX3 (X, -XZ) 10d, WX3WZ3 (
Y, -YZ) + d 4 C7l L) (WX4w
yr l'/4WXz))+ (Xi XZ)
((Ys YZ MZ3 (2522) I'yi
)+ (yi-yz)((zs zz)wx3(XS
Xz)Wzd+ (zi-Zz)((Xs-Xz)
ll'y3-(YS-Yz)Wx3)=tenb (d
3'iL-Moaning U□: Gochitsu□+d3 brother L-9 sudden bar work=
≦Shitsu.

-d・■Self ■g knee + d, crotch d-d3W
Wz, (X −X ) + d, Wx Wz Y −
Ya+ d4 (Wz3Wy 4 wy3 Wx4)
(Xi XZ) + d4(Wx, Wx4-1dz
, Wx4) (Yi-YZ) 1d 4 (wY3WIC4W
X3'JY4)CZj22))+(Xi Xz)
((Y,,YZ)Wjh (ZS 22)Wy
i ) + (yi-yz) ((zs -22)WX3-
(xs −XZ)Wx3)+ (Zi-22)((X
i-Xz)Wya-(YS-YZ)WX3)= 10b (10d, (WzJy4-Wy3111z4)(X
i-XZ) + d, (wx, wx4-Wz, Wx,)(
Yi-YZ) 10d4(Wy, Wx4Wx3Wy, 5)(
Zl-2z))+ (Xi-XZ)((YS-YZ)
WX3- (zs -Z2)Wya)+(YI YZ
)((ZS Zz)Wx, (x5 xz)bi
z3)+ (Zi Zz)((L Xz)WX3(
YS YZ)WX3)・・・・・・・・・・・・・・・
...(153) Scalar product (Xw3Wx, +Yw3
Wy6+ZwJzs) = Find b from O.

b=(: (XjXi)((YS YZ)11'z
3(ZS zz)wy3)(YI YZ)((25
22) WX3 (XS Xz)Wx3)-(zi-
Z2) ((Xi -xz)wyi -('ys -yz
)wxa))/(10d4(WZ31+ly4 Wy3
WZ4) (Xi'-XZ) 10d, (Wx3Wz4-W
z3Wx, ) (Yi-YZ) 10d 4 (Wya Wx
4Vx3111>'* ) (ZI Zz ))=(
-1/d4)((pi-P,)・((P,-P2)XW
,))/((W, X W3)・(Pi P2))・
・・・・・・・・・・・・・・・・・・(154) If we calculate Pw, , 2w4 from the following equation using the b obtained here, they will become the target P7, P, becomes. Second
In the flowchart of FIG. 6, step 45 is referred to.

Expressed as a formula, it can be obtained in the following form.

P7=Ps,=b-ci,w,+p2=w,(-c
i3/d4) [(pi-p2)-((p5-p2) x
w,))/((W, X W3)・(Pi-P2))
+ P2”-WIPi-P21C(Pi-P2)・((
p5-p2)xw3))/(lP4-Psi(W4XW
3)・(Pi Pz)) 10P2・・(155) P g =P 14== b−d 4 W 4 + P
=.

=W4(-d4/d,)((Pi-P2)・CCPS-
P2)X W3))/((W4XW,)・(Pi −P
z))+Ps=-W4((Px-Pz)・
((p5-p2)xw,))/((W4XW3)・(P
i - P2)) 10 P5 (156) Next, to express the corresponding positions when equally divided on another pair of opposing sides P2P5, sides P, and P4, , find the length when each is divided into n equal parts. Please refer to FIG. In addition, in the flowchart of FIG. 26, step 46
See. The length when side P2P is divided into n equal parts is d
If the length when dividing xz, sides P, and P into n equal parts is d i4, d 13 and d 14 can be obtained by the following equations.

d,,=l P5-P21/n-4201)d
14=l p4- P31/n=(202)
Next, the unit vector W 1 going from P2 to P5
3. Unit vector W construction going from P3 to P4.

make. In the flowchart of FIG. 26, step 47
See. The component values of W□3 and W14 are expressed as Wl3 = (
IN)h3+Wyx3. IjZ□i) ””””’(
203) W1+ = (WX1* +Wyxo +WZ
14) ”'”'”' (204), then Wla
The component value of is 1=x, -X2...(205) wk
2 = Y, -Y2---(206)wk3 = Z,
-Z2-(207) Suni 4 = ((υk12 + 22 for enemy + 32 for υ)...
・・・・・・・・・(208) Su X, = υ Nishini 4
−−−−・=−(209)Wy, 3=tik2/
wk4 --(210) Wz□3== wk3/
wk4...=...(211) The component value of W14 is wk5 = X4-X,...=...(21
2) 6 to υ = Y4-Y3 (2
13) wk7 = 24-2. -(214)w
k8 = v'(tyk52+wk62+wk72)-
--(215)Wx14=5 to υ/8 to υ...
...(216) y□4 = Suni 6/Suni 8 ・
...(217) Wz14=: wk7/ wk
8...-(218).

Using d□3 and Wl3 obtained above, a point Pw 3 on side P2P5 with P2 as a reference can be obtained in the following form.

In the flowchart of FIG. 26, reference is made to step 48.

P, 3=b-dl, W, +P2...
(219) Also, using d14 and Wl4 obtained above, the point Pw0. on side P3P4 with P3 as the reference.
is obtained in the following form Pw14=b-d14W14+P3 ・・・・・・・・・
...(220) Here, line segments Pw□, Pw0. However, if we define the variable so that it passes through one point Pi on the given surface, we can obtain another desired line segment. In other words,
Line segment Pw1. Pw1. All you have to do is decide the conditions for Pi to sit on top. To do this, it is sufficient to introduce a coordinate system in which the origin is Pw, and the reference axis is the direction from Pw□3 to P-□4, and the determination is made based on the distance of Pi from the reference axis. First, the origin is Pi,, =b-d13W□3+P2...
(221) Next, the reference axis is W, =Pw14-Pw, 3-...-=(222)=
b(d14W□, -dwork, Wwork,) 10p3-p2 To judge the degree of separation, an axis perpendicular to the reference axis is required, and if this axis is Wl6, then Wl, Haggi procedure. obtain.

Wig=W1. XW□、・・・・・・・・・(
223) Here, Wl,, Wl, is Wlg = (Wxxs+Wy1s+'Z□s) ・'
=□=”(224)W□c =(Wx1s+WyxsJ
Zzs) """"' (225), first, W
, 5 is obtained in the following form.

Wx,, = b (d L4wx14 d tJ
Xii) + (X3 X2)...(226) Wy□s = b (d x4wy14d tiIl
lyt3)+(Y3 Y2)・・・・・・・・・(22
7) bZxs = b (d 14”zi4 d 1J
Zta)+ (Zl Z2)・・・・・・・・・(2
2g) Next, each component value of W and G is obtained in the following form.

Wx,, =J'xs111Z1i WZxsl'1
y1a ”'”'”' (229) 11'yH;
=WZxsWX□i Wx,sWz,, ”'”'
”' (230) Wz16=l'XxsWy1311
')'xsl'X1z ”'”' ”・(231)W
Once □ is obtained, b can be found using the following equation.

(Pi−Pw, 3)·W, s=O---(232) Here, the values of each component of W5 are as follows.

Wx1s=Wy□JZz3WZzsWy, 3” (
b(dxJyx+-su,■,)”(Y3Yz))WZx
i(b (d 14wz14-A1 autumn,)+(za
zz))uyxa"b' d 14(Il'yx4
wZ13WZt41'yxz )+(Y3 Yz)W
Zt3(Z3L)Wy□:+”'(233)'dY1b
= WZxsWXt3Wx1sWz1゜” (b
(d i4wZi.-d Wz ) + (Z,
-22)) Wx,.

(b (d 14vx14 □) + (Xl Xz)
)WZzr= b °d, 4(WzzzWx□, −
Wx14Wz,, )+ (Zl −Z2)lilx,
, -(Xl-Xz)WZ13=-(234)'Zxr
, = 'x1sl'y13'JYxS'JX□3
” (b(dtJXt4m)+(Xi X2))'
! +'x3(b (d IJYxq-Ayotsujiyu)+
(Yr Yz))WXti= b ” d x4(W
X1+I'yx3'JYxj'Xx3)+(Xl L
)Wy□3(Yz Yz)Wx, 3''' (235)Here, if we create the scalar product (Pi-Pw□3)·W, we get the following.

Pi = (Xi, Yi, Zi) - (236) Pw, 3
=b−dl,W,,+P2 −・−・−−−(237
) First, (Pi PWzi) = (Xwt3+YWx3+ZWt
z)""""' (238), each component value is as follows.

Xw13=Xi-b-dl, Wxl, -X2-(239
) Ysu□, = Yi-b-d 工, enemyy□3-Y2...
・・・・・・(240)Zi+,,=Zi-b-d,3
Wz, 3-Z2・-・-・(241) Here, (Pi-Psu, 3)・W-work.

=XII□1Wx16+Ytnt311ye+Zwi3
Since Wzc -----(242), each term on the right side is expressed as: Xenemy□3Wx.

= (Xx b-d, 31ux13Xz) (b'dtq
(WyxJZx3'JZxJ'/x3) ten (YJ Y
z)uZl) (Z3Zz)I'y1a)...
...(243) =b2(d□ad14(WXtiWZxaWy141N
X□il'yxa'Z1j)+b (d 11]'x
l 3WZz3 (Yz Yz) + d 13”
xi31'Yx.

(23Zz)+d, 4C'JZx3'JY14WyzJ
Z□J(Xl-X2)) +(Xi X2)((Yl Yz)Wz13- (
Z3-Z2)Wy□3)Ysu□3sy16 = (Yi-b-d□3suy□a Yz) (b' d
14 (WZ, JXtzWx14Wz□3)+ (Z
3 Zz)WXz3(Xl -Xz)WZ□a)・・
・・・・・・・・・(244) =b2(di3dt4(WXxxWY13WZx4Wy
1311'Z+3WX□*)) 10b (d 1:+I'
+h:+l'yxz ('z3z2) tend1i'yxa
'Zia(X3X2) +d14(YI Yz)(W
xt:1Wzt4'JzxglIlx□4)) +(Yi Yz)((Z3 Zz)WXti (
L
(WXt4WyzaWytJXxa)+(L Xz)
W3h3(Yz Yz)WXi3)・・・・・・・・・
・(245) "b"(dxidt4(WZt31'ytJX14-
riZxxll'Xt3WytJ)+b (d 1Jy
xy'dZ□a (Xl X2)+ d 13”xl
ffw”43(Yz Yz)+d □*(ZI
Zz) (1'lXt*l'yxa WytJXt3)
)+ (ZI Z2)((Xl Xz)Wyii
(Yx Yz) WX13) Here, the scalar product (X11zJXtG+YVz311'yts
+ZW□3WZB) When you create ヲ, Xw, 3Wx1. +Yt+,,Jy,, +ZtI+
,,Wz1゜=b'' (d 13 d 14(W
XxaWZz3Wy□q WXtiWytzWZxJ
)+ b2(dzadx4(WXt3WytaWZz+
W3'z3WZ1iWXxJ)Jub"(dxad1
4 (WZt2111yz3WXz4-wz13suXta
W3'□J)+b (d 13vx13wzL3 (
Y3Y2) 10d □aWXxxWyta (Z3 22
)+d□4(WZx3wyx4Wy13WZt4)(X
i Xz)) 十b (d □atl'XzaWyzz
(23Z2)+ d 13wyziWZz3(Xl
X2) 14 (Yx Yz) (1'Xx3W
Zz4WZzal'xti))+b (d 13'1
lYty'Zxy (Xl X2)+ d 1. Wx
□, Wzlj (Yz-Yz) + dt4 (ZI Zz)
(WXxJyt311'ytJXxi))+ (Xi
X2) ((Yz Yz) WZ13 (Z3 Zz)
Wyx:+)+ (YI Yz)((Z3Zz)WX
t3(Xl L)Wzla)+(ZI Z2)((
L Xz)Wy13(Yz Yz)WXxa)・・
(246) From this, b can be obtained using the following formula.

b ” (-(xi-X2)((Yz-Yz)Wz13
- (2G-Zz)Wyza)(YI Yz)((Z
3 Z2)ll'X13 (Xl L)l'Zt
a)-(zz-zz)((x3-L)Wyzi-(Y
z-yz)"'x13))/(+d14(WZzaWy
x4WyzJZz4) (Xi Xz)+ d 1q
(WXtJZ14Wz13Wx14) (YjYz)+d
□4(Wyx3WXx4WXx:+wyz4)(ZI
Z2))=(1/d14)((Pi P2)・((P
3 P2)XWzi):1/((W14XW□3)・(
Pj P2))・・・・・・・・・(247) Using b obtained here, Pw 3,
If we find P wo, they are the objectives P5 and P8, respectively.
becomes. That is, in the flowchart of FIG. 26,
See step 49.

P6=Psu, 3=b-d□3wi,” P2=W 3
(-d engineering＞/d, 4)((Pi-P2)・((Pl-P
2) Xwis))/((Wt<xwtx)'(pi P
2))+P2=-W□31P-P21((PL-P2)
・((p3-pz)xwta)]/(IF5-P, 1(
W14XW Engineering, )・(Pi −p2))+ p2...
...(248) Pa"Pwz4=b-di4W,,+P3=Wl,
(-d 1. / d 4) [(P i-P 2)・((
P3-P2) x W 3)) / ((Wl4
□, )・(Pi−P2))+P.

=-Wyo4 ((P i-P 2)・((P3-P2)
×w13))/((Wl4 x Wla)・(p
j-p2)) 10p3... (249) Now, from the results obtained above, output the line segment connecting P7 and P9 and the line segment connecting PG and PIl as broken lines. Bye. In the flowchart of FIG. 26, reference is made to step 50.

Next, as shown in FIG. 14, an example in which the surface is a parallelogram in three-dimensional space will be described.

In addition, the flowchart of FIG. 27 is also used for explanation. In this case, as shown in FIG. 14, an image line segment is added as a dashed line from one point on the surface toward the sides constituting the surface, the direction of which is parallel to the side adjacent to the direction. This concept is also effective when the surface area is a triangle, for example. An example of a triangular shape is shown in FIG. In the case of a triangle, one point on the surface from each vertex of the triangle is shown in Figure 16.
It is conceivable to draw a broken line at two points, or to draw a broken line in the form of a perpendicular line drawn from one point on the surface to each side as shown in FIG. However, when we look at them as three-dimensional shapes, in the former it is easy to see one point on the surface as the vertex of a tetrahedron, and in the latter it is easy to see that a chain is stretched from one point on the surface to each edge. . In contrast, in the case of Figure 15, the line parallel to the broken line appears to be emphasized, so
It is easy to see two points as if they were on a surface. In this case, the information given includes the following (see FIG. 14). In the flowchart of FIG. 27, step 61 is referred to.

As one point on the surface, Pi = (Xi, Yi, Zi) = (300) As the coordinate value of the vertex of the parallelogram, P2 = (X,,Y,,Z2) -=-・-(301)
P, = (X3.Y,, 23) ・−・・・・・−・(
302) P4=(X4.Y,,Z4) --・----
...(303)Ps=(Xs+YsyZs) -...
-... (304) Also, as a result, for example, as shown in FIG. 14, there are two end points P6, P7, pi, and p of the added broken line. They are P+;=(Xs+Ys+Zs) ・−・・==−(3
05) P7=(X7.Y7.Z,) ・・・---・
・-(306) pH=(X,,YIl,Z,) ・-
---・-・(307)p,=(x,,y,,z,)
−・−−−−−−(308).

First, the unit vector U□ directing from P2 to P3,
Create a unit vector V moving from P2 to Ps. In the flowchart of FIG. 27, reference is made to step 62. The component values of Ul and ■□ are expressed as Ux” (LJXttlJ7x
+tlZx) −−−−−−−−−(309)■□=
(VXt+Vy engineering, Yz engineering) ・・・・・・・・・(31
0), then wk1 =
) 3=Z, -Zz to the police ・・・・・・・・・(
313) wk4 = v'(wk12+wk2" +
wk3”)---(314)lx□=wkl/wk
4 −−・・・−・(31s)Uy□=Suni2/w
k4・・・・・・・・・(316) Uz□= wk3
/ tllk4 − ... − (317) wk5=
X, -X2-- ... (318) wk6 = Y,
−Y2・= −・・・(319)wk7=Z, −
Z2・= --(320) wk8=v'(tgk5"
+wk62+wk72)−・−−−−−−(321
)Vx1=tgk5/wk8 ・−・−=−
(322) Vy, = wk6/wk8 ---
(323) Vz□=wk7/wk8 ・−・・
...-(:324).

Next, P z ” (X2.Yz, Zz) is the starting point, P
Direction vector W with i=(Xi, YitZi) as the end point
, find. In the flowchart of FIG. 27, reference is made to step 63.

If W3=(Wx,, Vy3.Wx3) - (325), then Wx3= Xi -X2--- (326)Wy, :
Yi −Y2 ・・・・・・・・・(327)
1iz3=Zi -Z2--- (328).

Here, the direction vector W3 is defined as the unit vector U□,■
When expressed as □, it becomes the following form. This is because W, U, and vl exist on the same plane.

W3=c1U1+c2V1...--4329) These 01 and 02 are found here. In the flowchart of FIG. 27, reference is made to step 64.

If they are expressed as component values, they will be in the following form.

Xl-X2= c 1Ux1+ c 2Vx, ---
(330) Yi-Y2=c, Uy□c2Vy,
・---(331)Zi-X2= cmUz□+c2
Vz16-(332) If we calculate 01 and 02 from these three equations, we get P6, P7, P, 1. P is obtained in the following form.

P, = c2V + P2 - (333) P, = C, U1 + P2 ... = - (334) P8
=c2■, 10P 3...-(335)P, =
c1U1+P5 - (336) This C engineering and 02 are the above formula (330) (331) (
332), but since there are two variables, two linear equations are used.

Xl-X2= c 1Ux, + c 2Vx□+++
+++ ... (337) Yi-Y2 = CxUy
x + c zVyt −= =・−・(338)vy
□(337) −Vx□(33g) is created, Vy,
(Xj-X2)=Vy1c 1Ux□+Vy, c
2Vx□VX Yi-Y =Vx c U +V
x c VVyl(Xi −X2) −Vxl(Yi
-Y2)” (VyxUXx VxlUyt) C□
c □= (Vyx (Xi X2 ) VX□(
YI Y2 ))/(VyiUxx-VX1U!/□
)・−・・・・−・(339) Substituting this into equation (337) to obtain c2, c z ”” ((Xi
X2) c 1Ux□)/Vxx= (Xi-X
2)/VLI -01(lx□/Vx□)=(Xi
X2)/Vx1 (Vyi(Xx X2) Vxl(YiYz))(
Uxt/VX□)/(VyllJxl vxtuyJ = (XjXz) (VyxUxx VxtUy□)/(
VXt (VyxUXz Vx4U3'z)) - (Vy
t (xi −X2) −Vx, (Yi −Y2)
) (UXl / Vxl)Vxt/ ((VyxUx
, -Vx□Uy1)Vx, )= (Vx, (Yi
Y2) UXlVxtUyx (Xx Xz))/((
VyzUXt VXxUyt)VXJ-'-cz=(
uxl(yi-yz)-uyx(xi-xz))/((
VytUxt−VxzUyt))−−・−・−・(34
0) Using the values of ci and 02 obtained here, the above formula (33
3) % formula % P8, P9 are obtained. In the flowchart of FIG. 27, reference is made to step 65.

After that, by creating a line segment connecting P6 and pH and a line segment connecting P7 and P9 as broken lines, the desired line display can be obtained. Second
In the flowchart of FIG. 7, reference is made to step 66.

The last example shown is a case where the surface is one spherical surface in three-dimensional space. An example is shown in FIG. Therefore, one point on the surface is
It is on the spherical surface. In addition, the flowchart in FIG. 28 will be used for explanation. In this case, a circle centered on the center of the sphere and passing through one point on a given surface is used as a characteristic line representing the surface of the surface. In this case, the information given includes the following: In the flowchart of FIG. 28, step 71 is referred to.

As one point on the surface, Pi=(Xi, Yi, Zi) ・−・−・・・−=−
・(300) As the center of the sphere, Pc=(Xc, Yc, Zc) ・−・・−−−−−−
(301) As the radius of the sphere, r ・・・・・・・・・・・・・・・(302) As the line of sight direction, k = (Kx, Ky, Kz) −−−−−・・・(3
03) As the position of the line of sight on the screen, P1=(Xl, Y,, Zl) -----=-・-・=
- (304). When there is no edge that represents the shape of the surface, such as a spherical surface, it is necessary to add characteristic lines to the part that can be seen from the viewing direction, so the viewing direction is used to obtain the characteristic lines.

First, the axis (
(hereinafter referred to as the reference axis). The reference axis here is
Let us give it as a vector Q pointing in a specified direction from the center of the sphere.

Qt = (Qx□, QylyQzt)...--
・－－－－－(305) Here Q□1=1 ・・・・・・・・・・・・・・・(30
6). At this time, a point P on the spherical surface above the reference axis Q
r□ is obtained in the following form.

Pr, = r Q, 10Pc・・・・・・・・・・・・
(307) Using point Pr1 on this spherical surface, create a characteristic line representing the surface of the surface in the form described below. First, the first characteristic line is defined by one point Pi on a given surface,
Create a circle that passes through the point Pr1 on the spherical surface obtained above and whose center is the center Pc of the sphere. Also, another characteristic line is
Create a circle that passes through one point Pi on the given surface and is perpendicular to the first characteristic line that has already been created.

In addition, the graphic processing device used here as an example is 3
When displaying a dimensional figure, it is assumed that the figure processing device has a function of displaying the two axes of the reference coordinate system so that they can be seen upward.

Here, the reference axis Q□ is made so as to stand upright in front.

To do this, it is first necessary to create the reference axis Q1 at a position rotated 90 degrees from the opposite direction (-k) to the direction of one line of sight in the Z-axis direction of the reference coordinate system of the graphic processing device.

First, from (-k) to the right side when looking at the two axes in front, (-k)
An axis B that is perpendicular to the Z axis and the Z axis is created. In the flowchart of FIG. 28, reference is made to step 72.

B = (Bx, By, Bz) -----=・(308
), B can be obtained by the following vector product.

B = (0,0,1) x (-k)...
(309) Expressed as component values, it is as follows.

Bx=-04z+l4y=Ky ・・・・・・・・・・・・・・・
(3], 0) By=-14x+04z = -Kx ・・・・・・・・・・・・・・・
・(311) Bz=-04y+04x 20...Yamakawa...(312) Here, B□, which is a unit vector of B, is obtained.

The component value is J = (Bxt, By,, Bzt) town - old... = - (3
13), B1 can be obtained in the following form. In the flowchart of FIG. 28, step 73 is referred to.

3) (1= Ky/v'(Ky2+Kx2) Byx=
Kx/V7″(Ky2+Kx2)Bz=0 Next, the desired reference axis Q□ is obtained by the following vector product. In the flowchart of FIG. 28, refer to step 74.

Q, =(-k)xB□・・・・・・・・・Town・(31
4) The component values are as follows.

Qx□=-Ky-Bz1+Kz-By.

= 10Kz-By1 = -KxKz/ v'(Ky" + Kx") -
・Old...Town...(315)Qyx =-Kz-Bx□+
Kx−Bz□” −Kz−Bxl = −KzKy/ J (Ky2+Kx2) −−・・
・Scream...(316)Qz1=-Kx-By, +Ky-B
x1= + (KxKx + KyKy) / v'
(Ky2+Kx2)"v' (Ky2+Kx2)
-----(317) One reference axis is now set. FIG. 19 shows an example of display using this reference axis. In this case, as shown in the figure,
The first circle set (one of the dashed circles in the figure) becomes an upright circle. This is because the reference axis is upright.

Now, FIG. 20 shows an example in which this is slightly tilted. For example, the hand j when rotated to the right by an angle θ
@ is shown below.

First, when the reference axis Q□ is rotated to the right by an apparent angle θ in the direction of the Bi axis, the component value in the Q upward direction is CO5θ, and the component value in the B1 direction is sinθ (see FIG. 21).

Therefore, Q2, which visually indicates the direction when the reference axis Q□ is rotated in the B1 direction by an angle θ, can be obtained by the following equation. 28th
In the illustrated flowchart, reference is made to step 75.

Qz=(cosθ)Qi+(sinθ) B, −”
・−・−・−−−− (318) If expressed as a component value, Q2 = (Ql2 +Qyz, QI2)−=−・
----=-(319) as follows.

Qx2: (-KxKz cosθ+Ky sinθ)
/((Ky2+Kx2)・・・・・・・・・・・・・・・
・(320)Qy2= (−KyKz cosθ−Kx
sinθ) / J (Ky2+Kx”)...
・・・・・・・・・・・・(321)Qz2= (co
sθ)v' (Ky2+ Kx2) ------(
322) Furthermore, there are cases where it is desired to rotate the reference axis visually toward the front. The configuration at that time is shown in FIG.

In this case, the form of shaft rotation is shown in FIG. in this case,
That is, Q3, which indicates the direction when the reference axis Q2 is rotated by an angle α, visually toward the front, is obtained by the following equation. In the flowchart of FIG. 28, reference is made to step 76.

Q3=Q, (cos a) (cosθ)+
(-k) (sin a) (cos θ)...
・・・・・・・・・(323) Expressed in component values, Q3 = (QI3 , Qx3 tQZ3)"'・"
・・・−(324) is as follows.

Qx, = (cos a ) (cos θ) (-KxK
z/J (Ky2+Kx2))+ (sin a)
(cosθ)(-Kx) ---・--(325)Qx
3 = (cos a ) (cos θ) (KzKy/
v' (Ky2+Kx2))+ (sin a) (
cos θ) (−Ky) −−−−=・−・−・(32
6) Qz3= (cos a ) (cos θ) (v'
(Ky2+ KX2))+ (sin a)(cos
e ) (-Kz) −−=−・=−(327) Next,
The straight line in the direction of the reference axis Q1 obtained above passes through the point Pr1 where it intersects with the spherical surface and one point Pi on the given surface, and a circle C whose center is the center Pc of the sphere is created. Point Pj1 on the circumference in this case is obtained by the following procedure. Further, the calculation method is, for example, that Pjx is calculated by giving an angle β in the direction from Pi to Pr1. Note that the same applies to the cases of Q2 and Q, in which case Ql may be read as Q2 and Q3, respectively.

First, vector (Pc, Pi) is transformed into vector (Pc, Pr
An axis Q4 is created at a position rotated 90 degrees in the direction of ). Second
In the flowchart of FIG. 8, step 77 is referred to.

For this purpose, if the unit vector of the vector (Pc, Pi) is Qi, then Qi=Pi−Pc
・・・・・・・・・・・・・・・(328) Vector (P
Since the unit vector of (c, Pr) is Ql, the axis B2 orthogonal to Qi and Q□ can be obtained by the following vector product.

B 2 ” Q x X Q□・・・・・・・・・・・・
(329) Using this, the axis Q4 to be sought can be obtained by the following vector product.

Q, =・B2XQj/ B2×Qi ・・・・・・・・・
・・・・・・・・・(330) Here, Qi is the first axis, Q4
Considering a plane coordinate system having Q as the second axis, the position rotated by an angle β□ from Qi in the Q4 direction has the following axis component value.

First axis component value = cosβ1 ・・・・・・・・・・・・
...(331) Second axis component value = sinβ1
(332) Therefore, the desired point Pjz on the circumference can be obtained by the following equation. In the flowchart of FIG. 28, reference is made to step 78.

P j, = r (cos β1) Qi+ r (si
nβ1)Q4+PC・・・・・・・・・・・・・・・(
333) After that, the necessary value can be output using Pj□.

Next, the circle C obtained above, that is, the straight line in the direction of the reference axis Q1, passes through the point Pr□ where it intersects with the spherical surface and one point Pi on the given surface, and the center is a circle with the center Pc of the sphere. It is perpendicular to C1, passes through one point Pi on the surface, and the center is the center of the sphere Pc
Create a circle C2. In this case, Qi obtained above
Consider a plane coordinate system in which Q5 is the first axis and B2 is a unit vector, and Q5 is the second axis. First, Q is obtained by the following formula. In the flowchart of FIG. 28, reference is made to step 79. The same applies to Q2 and Q3, and in that case, Ql may be read as Q2 and Q3, respectively.

QS” (QI X Ql) / l QI X Qll
------- (334) Here, Qi is the first axis, Q5
In a planar coordinate system having Q as the second axis, a position rotated by an angle β2 from Qi in the Q5 direction has the following axis component value.

First axis component value = cosβ2 ・・・・・・・・・・・・
...(335) Second axis component value = sinβ2
(336) Therefore, the desired point Pj2 on the circumference can be obtained by the following equation. In the flowchart of FIG. 28, reference is made to step 8o.

P j2= r (cos β,)Qi+ r (sin
β2) After QS+Pc, use PJz to calculate the required value by 8
All you have to do is force yourself.

[Effects of the invention] It becomes possible to visually identify whether a displayed point is on a surface or not.

BRIEF DESCRIPTION OF THE DRAWINGS FIG. 1 is a diagram showing an example of the configuration and data flow of a graphic processing device to which the present invention relates. FIG. 2 is a diagram showing an example of the form displayed on the screen. FIG. 3 is a diagram showing an example of one surface existing in three-dimensional space. FIG. 4 is a diagram showing an example in which one point on the screen is given as information specifying one point on the surface. FIG. 5 is a diagram showing an example of displaying a surface on a screen using the viewing direction. FIG. 6 is a diagram showing an example of determining one point on a surface (the point where a straight line passing through a point on the screen and parallel to the viewing direction intersects with a given surface). Figure 7 shows the characteristic line representing the surface of a surface.
The figure which shows the example which added and displayed the part regarding one point. FIG. 8 is a diagram showing an example in which a characteristic line representing the surface of the surface is added to and displayed at one point on a surface existing on a spherical surface. FIG. 9 is a diagram showing an example in which a characteristic line representing the surface of a surface is added to one point on a surface existing on a general quadrilateral. FIG. 10 is a diagram showing an example of information given in the case of a rectangle. FIG. 11 is a diagram showing an example of information obtained as a result in the case of a rectangle. FIG. 12 is a diagram showing an example of the concept in the case of a general quadrilateral. FIG. 13 is a diagram showing an example of the concept in the case of a general quadrilateral. FIG. 14 is a diagram showing variable assignment in an example of a parallelogram. FIG. 15 is a diagram showing an example of applying the concept of a parallelogram to a triangle. FIG. 16 is a diagram showing an example of an unfavorable concept when applied to a triangle. FIG. 17 is a diagram showing an example of an unfavorable way of thinking when applied to a triangle. FIG. 18 is a diagram showing an example in which a characteristic line representing the surface of a surface is added to one point on a surface existing on a spherical surface. FIG. 19 is a diagram showing an example in which when a circle is used to add a characteristic line representing the surface of a surface, the reference axis is set so that the circle stands upright. FIG. 20 is a diagram showing an example in which when a circle is added to a characteristic line representing one surface, the reference axis is set so that the circle tilts left and right. FIG. 21 is an explanatory diagram of the idea of tilting a circle in three-dimensional space to the left and right at its apparent position. FIG. 22 is a diagram showing an example in which when a circle is used to add a characteristic line representing the surface of a surface, the reference axis is set so that the circle is inclined toward the front or toward the rear. FIG. 23 is an explanatory diagram of the idea of tilting a circle in three-dimensional space forward and backward at an apparent position. FIG. 24 is a flowchart of an example of the operation of the first means 1. FIG. 25 is a flowchart in the case of a rectangle. FIG. 26 is a flowchart for a general quadrilateral. FIG. 27 is a flowchart in the case of a parallelogram. FIG. 28 is a flowchart in the case of a spherical surface. Explanation of symbols 1...first means, 2...second means, 3...processing section, 4...input section, 5...output section. 20th figure G leg 7■ 19th figure 3■ Gisen〇 screen j/l direct, 9th fossa 1? Kuni 19th 4 ■ 25th ward 2b exit (Mu1) 27th Senwa gJ (Zon n2)

Claims (1)

[Claims] 1. In a graphic processing device having at least an input section and a display section, information indicating a surface existing in a three-dimensional space inputted from the input section, information specifying a point on the surface, and the three-dimensional space are provided. Obtain points on the surface based on information indicating the direction in which the surface is viewed,
A graphic processing apparatus characterized in that a characteristic line representing the surface of a surface is obtained, added to a portion related to a point on the surface, and displayed on the display section. 2. In a graphic processing device having at least an input section and a display section, information input from the input section indicating a surface existing in a three-dimensional space, information specifying a point on the surface, and information indicating a surface in the three-dimensional space are inputted from the input section. Obtain information indicating the viewing direction, obtain a point on the surface from information indicating a surface existing in the three-dimensional space, and information specifying a point on the surface, and connect the point on the surface and the three-dimensional space. A characteristic line representing the surface of the surface is obtained from information indicating the surface existing in the space and, if necessary, information indicating the direction in which the surface of the three-dimensional space is viewed, and added to the part related to the point on the surface. A graphics processing device, wherein information indicating a direction in which a surface of the three-dimensional space is viewed is displayed on the display unit.