Intersection of regular solids
When two solids Interpenetrate, a line of intersection is formed. H is sometimes necessary to know the exact shape of this line, usually so that an accurate development of either or both of the solids can be drawn. This chapter shows the lines of intersection formed when some of the simpler geometric solids interpenetrate
Two dissimilar equare prleme meeting at right angles (Fig. 12/1)
The E E. shows where corners 1 and 3 meet the larger prism and these are projected across to the F.E. The plan shows where comers 2 and 4 meet the larger prism end this Is projected up to the F.E.
Two dieeimilar square prisma meeting at an angle
The F.E. shows where corners 1 end 3 meet the larger prism. The plan shows where corners 2 and 4 meet the larger pdsm and this is projected down to the F.E.
3R0 ANGLE PROJECTION 2
3R0 ANGLE PROJECTION 2
1ST ANGLE PROJECTION
3RD ANGLE PROJECTION A hexagonal prism masting s square prism at right
3RD ANGLE PROJECTION A hexagonal prism masting s square prism at right
Two dissimilar hexagonal prisms meeting at an angle (F»g. 12/4)
The F.E shows where comers 3 and 6 meet the larger prism The plan shows where corners 1,2.4 and 5 meet the Larger prism and these are projected up to the F E.
12/4
ANGLE PROJECTION
12/4
ANGLE PROJECTION
A hexagonal prism meeting an octagonal prism at en angle, their centre« not being in the aame vertical plana (Rg. 12/6)
The F.E. shows where corners 3 end 6 meet the octagonal prism. The plan shows where comers 1, 2. 4 and 5 meet the octagonal prism and these are projected down to the Ft
The sides of the hexagonal prism between corners 34 end 56 meet two sides of the octagonal prism. The change of shape occurs at points a and b. The position of a and b on the F.E. (and then across to the E.E.) is found by projecting down to the F.E. via the end of the hexagonal prism (follow the arrows). The intersection on the F.E. can then be completed.
A squire prism msstlng a squsre pyramid at right anglaa (Fig. 12/6)
Tha E.E. shows where corners 1 and 3 meet the pyramid. These are projected across to the F.E.
Corners 2 and 4 are not quite so obvious. The pictorial view shows how these comers meet the pyramid If the pyramid was cut across XX. the section of the pyramid resulting would be square, and points 2 and 4 would lie on this square. It isn't necessary to make a complete, shaded section on your drawing but it is necessary to draw the square on the plan Since points 2 and 4 lie on this square it is simple to find their exact position Project comers 2 and 4 from the E.E. onto the plan. The points where these projectors meet the square are the exact positions of the intersections of corners 2 and 4 with the pyramid.
A square pyramid and • hexagonal prism meeting
The F.E. shows where corners 1 end 4 meet the pyremid.
Comers 2 and A lie on the same plane XX If this plane is marfced on the plan view of the pyramid (follow the arrows) it results in the line XXX Corners 2 and 8 lie on this plane; their exact positions are as shown
Corners 3 end 5 lie on the same plane YY. On the plan view this plane is seen as the tine YYY (follow the arrows) Comers 3 and 5 lie on this plane; their exect positions are as shown.
Two dieeimilar cylinder« meeting at right angle« (Fig. 12/8)
The smaller cylinder is divided into 12 equal sectors on the F E. and on the plan (the E.E. shows how these ere arranged round the cylinder).
The plan shows where theee sectors meet the lerger cylinder and these intersections are projected down to the F.E. to meet their corresponding sector at 12". 3'. etc.
3RD ANGLE PROJECTION
)7  
[S? 1 
Two dissimilar cylinder« meeting at an angle (Fig 12/9)
The method is identical with that of the last problem. The smaller cylinder is divided into 12 equal sectors on the F.E. and on the plan
The plan shows where these sectors meet the larger cylinder and these intersections are projected up to the F.E. to meet their conesponding sectors at 1*. 2'. 3\ etc.
Two dissimilar cylinder« meeting at an angle, their centra« not being In the same vertical plane (Fig. 12/10)
Onca again, the method is identical with that of the previous example. The smaller cylinder is divided into 12 equal sectors on the F.E and on the plan
The plan shows wher« the sectors meet the larger cylinder and these intersections are projected down to the F.E to meet their corresponding sectors at 1'. 2', 3'. etc.
A cylinder meeting a square pyramid at right angles
The F.E. shows where points 1 and 7 meet the pyramid and these are projected down to the plan.
Consider the position of point 2. Since the cylinder and the pyramid interpenetrate, point 2 lies on both the cylinder and the pyramid. Its position on the cylinder is seen fairly easily. On the F.E. it lies on the line marked 2.12 and on the plan it lies on the line marked 2.6. Its position on the pyramid is not quite so obvious. Imagine, on the F.E. that the part of the pyramid that is above the line 2.12 was removed. The section that resulted across the pyramid would be a square and point 2 would lie somewhere along the perimeter of that square It isn't necessary to construct a complete, shaded section across the pyramid at line 2.12 but the square that would result from such a section is constructed on the plan. In Fig. 12/11 this is marked as 'SQ 2.12'. Since point 2 lies somewhere along the line 2.6 (in the plan) then its exact position is at the intersection of the square and the line. This is shown on the plan as 2'.
Point 12' is the intersection of the same square and the line 8.12 (in the plan).
Thia process is repeated for each point in turn. When the intersection has been completed in the plan, it is a simple matter to project the points up on to the F.E. end draw the intersection there.
1ST ANGLE PROJECTION
1ST ANGLE PROJECTION
A cylinder meeting a square pyramid at an angle (Fig. 12/12)
The F.E. shows where points 1 and 7 meet the pyramid and these are projected down to the plan.
Consider the position of the point 2. In the F.E it lies somewhere along the line marked 2.12 whilst in the plan it lies on the line marked 6.2. If that part of the pyramid above the line. 2.12 in the F.E. was removed, the point 2 would lie on the perimeter of the resulting section. This perimeter can be drawn on the plan and. in Fig. 12/12, it is shown as the line marked 'SECT 2.12". Point 2 must lie on this line: it must also he on the line marked 6.2 and its exact position is the intersection of these two lines.
Point 12' is the intersection of the same section line and the line marked 8.12.
This process is repeated for each point in turn. When the plan is complete the intersection can be projected onto the other two views. For the sake of clarity, these projections are not shown.
1ST ANGLE
A cylinder meeting e hexagonal pyramid at an angle
Onca again lines are drawn on the plan which represent the perimeters of sections taken on the F.E. on lines 1; 2. 12; 3. 11. etc All the construction lines on Fig. 12/13 ere for finding these section perimeters.
The line of mterpe net ration. first drawn on the plan, is the intersection of the line 1.7 with section 1. line 2.6 with eection 2.12 (giving point 2*). line 3.5 with section 3.11 (giving point 3*), line 4 with section 4.10. etc.
When the intersection is complete on the plan, it can be projected onto the other two elevations For the sake of clanty, these projections are not shown
A cylinder meeting a cone, the cone enveloping the cylinder (Fig. 12/14)
The cylinder it divided into 12 equal sector» on the F.E. and on the plan.
Consider point 2. On the F.E. It lies somewhere along the line marked 2.12 whilst on the plan it lies on the line marked 2.6. If. on the F.E.. that part of the cone above the line 2.12 was removed, point 2 would lie somewhere on the perimeter of the resulting section. In this case, the section of the cone is a circle and the radius of that circle is easily projected up to the plan. In Fig. 12/14, the section is marked on the plan as 'SECT 2.12' and the exact position of point 2' is the intersection of that section and the line marked 2.6. Point 12' is the intersection of the same section and the line marked 12.8
This process is repeated for each point in turn When the plan is complete, the points can be projected down to the F.E.; this is not shown for cianty.
A cylinder and a cona. neither enveloping the other
The constructions ere exactly the tame at thote uted in the previout example with one tmail addition
The E.E. shows a point of tangency between the cylinder and the cone This point is projected across to the F.E. and op to the plan as shown
A cylinder and a con«, th« cylinder enveloping the con« (Fig. 12/16)
The construction required here is a modified version of the two previous ones. Instead of the cylinder being divided into 12 equal sectors, some of which would not be used, a number of points are selected on the E E. These are marked on the top part of the cylinder as a. b and c whilst the lower pan is marked 1.2.3 end 4.
As before the sections of the cone across each of these points are projected up to the plan from the F.E. Each point is then protected from the E E. to meet its corresponding section on the plan at a', b'. c*. 1'. 2', 3" and 4\
These points, are then protected down to the F.E. For the seke of clarity, this is not shown.
I 1ST ANGLE
PROJECTION
I 1ST ANGLE
Fig. 12/17
PROJECTION
A cylinder meeting a hemisphere (Fig. 12/18) The cylinder is divided into 12 equal sectors on the F.E. end on the plan.
Sections ere projected onto the plan from the F.E. The section planes are level with the lines 1 ; 2.12; 3.11 ; 4.10. etc. and these sections appear on the plan as circles.
On the plan, the sectors from the cylinders are projected across to meet their respective section at 1 2'. 3*. 4' etc.
When the interpénétration is complete on the plan, it can be projected up to the F.E. For the sake of clarity this construction is not shown.
Fig. 12/18
A cylinder meeting a cone, their centres not being in the same vertical plane (Fig. 12/17) Divide the cylinder into 12 equal sectors on the F.E and on the plan
Sections are projected from the F.E. to the plan, the section planes being level with lines 1; 2,12; 3.11; 4.10. etc. These sections appear on the plan as circles. On the plan the sectors from the cylinder are protected across to meet their respective section at points 1'. 2*. 3'. etc. The complete interpenetration can then be projected up to the F.E. For the sake of clanty. this construction is not shown.
1ST ANGLE PROJECTION
Fig. 12/17
A cylinder meeting a hemisphere (Fig. 12/19) The solution is exactly the same as the last example except the sections are projected onto tha E.E. and not the plan.
FILLET CURVES A sudden change of shape in any loadbearing component produces a stress centre, that is. an area that is more highly stressed than the rest of the component and therefore more liable to fracture under load. To avoid these sharp comers, filler radii are used These radii allow the stress to be distributed mora evenly, making the component stronger
1ST ANGLE PROJECTION
FILLET RADIUS
1ST ANGLE PROJECTION
FILLET RADIUS
Sometimes, parts of these fillet radii are removed and a curve of intersection results. Fig. 12/20 shows an example of this.
Sections are teken on the F.E. These appear on the plan as circles. The points where these sections 'run off the plan can easily be seen (at 1. 2. 3 and 4) and they are projected up to the F.E. to taeet their respective sections
3RD ANGLE PROJECTION
Fig. 12/21 shows how a fillet radius could be used on the end of a ring spanner
Sections ere taken on the F.E. end projected up to the plan (for the sake of clarity, the projection lines for the sections are not shown). The points where these sections run off the plan can easily be seen and these points (1. 2. 3. 4 etc.) are projected down to the F.E to meet their respectrve sections in 12'. 3". A', etc.
3RD ANGLE PROJECTION
Fig. 12/21
To find the true length of e line that Is not parallel to any of the principal planes and to find the angle that the line makes with the F.V.P. (Fig. 13/2) The line is AB. On the F.V.P. it is seen as ab and on the H Pas a,6,.
One end ol the line A is kept stationary whilst B is swung round so that AB is parallel to the H P. B is now at B' and on the F.V.P. b is now at b'. Since the line is parallel to the H.P. it will project its true length onto the H.P. This is shown as axbt. Notice that b, and b, are the same distance from the line XY.
Since AB' (and ab') are parallel to the H P., the angle that AB makes with the F.V.P. can be measured This is shown as 0.
INITIAL CONDITIONS (ABIS NOT PARALLEL TO EITHER PLANE)
ENO A IS KEPT STATIONARY AND B IS SWUNG ROUND SO THAT AB IS PARALLEL TO H.R
LINE IS PROJECTED FROM NEW POSITION TO SHOW TRUE LENGTH ON KP ANO THE ANGLE THAT LINE MAKES WITH F.V.P
To find the true length of a lina that ia not paral!®! to any of tha principal planea and to find the angla that the line mekei with the H.P. (Fig. 13/3) The line is AB. On the F.V.P. ¡t is seen as ab and on the H.P. as*./»,.
One end of the line B is kept stationary whilst A is swung round so that AB is parallel to the F.V.P. A is now at A' and. on the H.P.. a is now at #'. Since the kne is now parallel to the F.V.P. it will project its true length onto the F.V.P This is shown as «¿r. Notice that a, and o are the same distance from the line XY.
Since BA (and b,a') are parallel to the F.V.P., the angla that AB makes with the H.P. can be moasured. This is shown as
LENGTH AE
TRUE LENGTH TRUE LENGTH
Fig. 13/4 is an example of an application of the theory shown above. It shows how simple it is to apply this theory.
A pylon is supported by three hawsers. Given the plan and elevation of the hewsers. find their true lengths and the angle that they make with the ground.
In the plan, each hawser is swung round until it is parallel to the F.V.P. The new positions of the ends of the hawsers are projected up to the F.E. end joined to the pylon at A end B. This gives the true lengths and the angles.
TRUE LENGTH TRUE LENGTH
LENGTH AE
To find the traces of a straight line given the plan and elevation of tha line (Fig. 13/6) The line is AB. If the line is produced it will pass through both planes, giving traces T. and Tn.
»b is produced to meet the XY line. This intersection is projected down to meet a ,b, produced in T*.
»b is produced to meet the XY line. This intersection is projected down to meet a ,b, produced in T*.
To draw the alavatlon and plan of a Una AB given it* true length end the distances of the ends of the lina from tha principal planes, in this case a, and
1. Fix points a and a, at the given distances av and a» from the XY line. These are measured on a common perpendicular to XY.
2. Draw a line parallel to XY, distance b, from XY.
3. With centre a. radius equal to the true length AB. d'aw an arc to cut the line drawn parallel to XY in C.
4. From a, draw a line parallel to XY to meet a line from C drawn perpendicular to XY in 0.
5. Draw a line parallel to XY. distance b» from XY.
6. With centre *„ radius a.D, draw an arc to cut the line drawn parallel to XY in b,.
7. Draw a line from b „ perpendicular to XY to meet the line drawn parallel to XY through C in b sb is the elevation of lhe line, a J>, is the plan of the line.
To construct the plan of a line AB given the dratance of one end of the line from the XY line in the plan (a«), the true length of the line and the elevation (Fig. 13/7)
1. From b draw a line parallel to the XY line
2. With centre a, redius equal to the true length of the line AB. draw an ere to cut the parallel line in C.
3. From a, (given), draw a line parallel to the XY line to meet a line drawn from C perpendicular to XY in D
4. With centre *,. radius equal to *,D draw an arc to meet a line drawn from b perpendicular to XY in b,.
3RD ANGLE PROJECTION
3RD ANGLE PROJECTION
> 0 
— 
7 '  
Fig 13/6 ^"'■J 
1 J 
3RD ANGLE PROJECTION
3RD ANGLE PROJECTION
1ST ANGLE PROJECTION
3R0 ANGLE PROJECTION
GIVEN
ANGLE Fig. 13/9
Fig. 13/10
ALTERNATIVE SOLUTION
Y TRACES OF PLANE
1ST ANGLE PROJECTION
To construct tha alevation of a lina AB given tha plan of tha lina and tha angla that tha lina makee wlth tha horizontal plane (Fig. 13/9)
1. Draw tha plan and from ona and eract a perpendicular
2. From tha othar end of tha plan draw a lina at the angle given to intersect tha perpendicular in C.
3. Fromb, draw a line perpendicular to XY to meet XY in b. 4 From a, draw a line perpendicular to XY and mark off
XYto* equal toa.c ab ia the required elevahon An attemetive Solution it alaoshown.
ANGLE Fig. 13/9
3R0 ANGLE PROJECTION
GIVEN
Fig. 13/10
ALTERNATIVE SOLUTION
To draw the plen of a line AB given the élévation of the line and tha angle that the line makea with the vert Icel plane ( Fig 13/10)
This construction is very similar to the last one and can be followed from the instructions given for that exemple.
Y TRACES OF PLANE
THE INCLINED PLANE
Définition
An inclined plane is inclined to two of the principal planes and perpendicular to the third.
Fig. 13/11 shows a rectangular plane that is indined to the H.P. end the E.V.P. end is perpendicular to the F.V.P. Because it is perpendicular to the F.V.P.. the true angle between the inclined plane and the H P. can be measured on the F.V.P. This is the angle 4
PROJECTION OF PLANE Fig 13/11
PROJECTION OF PLANE
The top drawing shows the traces of the plane after rabatment. The bottom drawing shows the full projection of the plane. It should be obvious how the full projection is obtained if you are given the traces and told that the plane is rectangular
Fig. 13/12 shows a triangular piano inclined to the F.V.P. and the E.V.P.. and perpendicular to the H.P. Because it is perpendicular to the H.P.. the true angle between the inclined plane and the F.V.P. can be measured on the H P. This angle is $.
3RD ANGLE PROJECTION
Once again, it should be obvious how the full projection of the inclined plane it obtained if you are given the traces and told that the plane is triangular.
TRACES OF PLANE m*
Fig 13/12
PROJECTION OF PLANE
To find the true shape of en inclined plene If the inclined plane is swung round so that it is parallel to one of tho reference planes, the true shape can be protected In Fig 13/13. the plan of the plane. HT, is swung round to HT. The true shape of the plane can then be drawn on the F V.P.
3RD ANGLE PROJECTION
TRACES OF PLANE
Fig 13/15
Fig. 13/14
3RO ANGLE PROJECTION
SIDE A
Fig 13/15
PROJECTION OF PLANE
THE OBLIQUE PLANE
Definition
An oblique plane is a plane that is inclined to all of the principal planes
Fig. 13/15 shows a quadrilateral plane that is inclined to all throe principal planes.
3RD ANGLE PROJECTION
Fig. 13/14
Fig. 13/14 shows an example. An oblique, truncated, rectangular pyramid stands on its base. The problem is to find the true shape of sides A and B
In the F.E side A is swung upright and its vertical height is projected across to the E E. whero the true shape of side A can be drawn.
In the E.E. side B is swung upright and projected across to the F.E. where the true shape is drawn
TRACES OF PLANE
3RO ANGLE PROJECTION
SIDE A
3. Fig 3 shows the elevation of a line AB which has a true length of 100 mm End B of the line is 12 mm in front of the V P.; end A is also «n front of the V P. Draw the plan and elevation of this line and determine and indicate its V.T and H.T Measure, state and indicate the angle ol inclination of the line to the H P. Joint Matriculation Board
Fig 3
Fig 3
DIMENSIONS IN mm
DIMENSIONS IN mm
4. A line AB of true longth 88 mm lies in an auxiliary vertical plane which makes an angle of 30° with the vertical plane. The line is inclined at an angle of 45° to the horizontal, the point B being the lowest at a vertical distance of 12 mm above the honzontal plane and 12 mm in front of the vertical plane Draw the plan and elevation of AB and cleerly identify them In the drawing. Associated Examining Board
7. The projections of a tnangle RST are shown in Fig. 6 Determine the true shape of the tnangle. Associated Examining Board
:. The plan and elevation of two straight lines are given in Fig 7. Find tho true lengths of the lines, the true angle between them and the distance between A and C
Southern Universities' Joint Board
5. The plan of a line 82 mm long is shown in Fig. 4 The elevation of one end is at b'. Complete the elevation and measure the inclinations of the bne to the H P. and V P.
University of London School Examinations
i  
f 
Fig 4 DIMENSIONS N mm6. Fig 5 shows the plan of end A of a line AB. End A is m the plane H.T.V End B is in the H P The line AB is porpendicular to the plane H.T.V Draw the plan and elevation of AB. Joint Matriculation Board (H.V.T. = Horizontal end vertical traces) I Fig 8 shows the plan and elevation of a triangular lamina. Draw these two views and. by finding the true length of each side, draw the true shape of the lamina Measure and state the three angles to the nearest degree Oxford and Cambridge Schools Examination Board Fig 8 Fig 8 DIMENSIONS IN mm DIMENSIONS IN mm 10. At Fig. 9 are shown the elevation and plan of a triangle ABC. Determine the true shape and size of the triangle University of L on don School Examinations11. Fig. 10 shows two views of an oblique triangular pyramid, standing on its base Draw the given view together with an auxiliary view looking in the direction of arrow A which is perpendicular to BC Also draw the true shapes of the sides of the pyramid. Note: Omit dimensions but show hidden detail in all views. Scale : full size. Oxford Local Examinations (See Ch. 10 for information not in Ch. 13) Fig 10 Fig 10 
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how I draw the intersection between the vertical cylinder and constep by stepthe vertical cylinder and con n between7 years ago
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