_PLOTTING MECHANICAL MOTIONS._
[Illustration: Fig. 257.]
Let it be required to find how much motion an eccentric will give to its
rod, the distance from the centre of its bore to the centre of the
circumference, which is called the throw, being the distance from A to B
in Figure 257. Now as the eccentric is moved around by the shaft, it is
evident that the axis of its motion will be the axis A of the shaft.
Then from A as a centre, and with radius from A to C, we draw the dotted
circle D, and from E to F will be the amount of motion of the rod in the
direction of the arrow.
This becomes obvious if we suppose a lead pencil to be placed against
the eccentric at E, and suppose the eccentric to make half a revolution,
whereupon the pencil will be pushed out to F. If now we measure the
distance from E to F, we shall find it is just twice that from A to B.
We may find the amount of motion, however, in another way, as by
striking the dotted half circle G, showing the path of motion of B, the
diameter of this path of motion being the amount of lateral motion given
to the rod.
[Illustration: Fig. 258.]
In Figure 258 is a two arm lever fast upon the same axis or shaft, and
it is required to find how much a given amount of motion of the long arm
will move the short one. Suppose the distance the long arm moves is to
A. Then draw the line B from A to the axis of the shaft, and the line C
the centre line of the long arm. From the axis of the shaft as a centre,
draw the circle D, passing through the eye or centre E of the short arm.
Take the radius from F to G, and from E as a centre mark it on D as at
H, and H is where E will be when the long arm moves to A. We have here
simply decreased the motion in the same proportion as one arm is shorter
than the other. The principle involved is to take the motion of both
arms at an equal distance from their axis of motion, which is the axis
of the shaft S.
[Illustration: Fig. 259.]
In Figure 259 we have a case in which the end of a lever acts directly
upon a shoe. Now let it be required to find how much a given motion of
the lever will cause the shoe to slide along the line _x_; the point H
is here found precisely as before, and from it as a centre, the dotted
circle equal in diameter to the small circle at E is drawn from the
perimeter of the dotted circle, a dotted line is carried up and another
is carried up from the face of the shoe. The distance K between these
dotted lines is the amount of motion of the shoe.
In Figure 260 we have the same conditions as in Figure 259, but the
short arm has a roller acting against a larger roller R. The point H is
found as before. The amount of motion of R is the distance of K from J;
hence we may transfer this distance from the centre of R, producing the
point P, from which the new position may be marked by a dotted circle as
shown.
[Illustration: Fig. 260.]
In Figure 261 a link is introduced in place of the roller, and it is
required to find the amount of motion of rod R. The point H is found as
before, and then the length from centre to centre of link L is found,
and with this radius and from H as a centre the arc P is drawn, and
where P intersects the centre line J of R is the new position for the
eye or centre Q of R.
[Illustration: Fig. 261.]
In Figure 262 we have a case of a similar lever actuating a plunger in a
vertical line, it being required to find how much a given amount of
motion of the long arm will actuate the plunger. Suppose the long arm to
move to A, then draw the lines B C and the circle D. Take the radius or
distance F, G, and from E mark on D the arc H. Mark the centre line J of
the rod. Now take the length from E to I of the link, and from H as a
centre mark arc K, and at the intersection of K with J is where the eye
I will be when the long arm has moved to A.
[Illustration: Fig. 262.]
In Figure 263 are two levers upon their axles or shafts S and S'; arm A
is connected by a link to arm B, and arm C is connected direct to a rod
R. It is required to find the position of centre G of the rod eye when D
is in position E, and when it is also in position F. Now the points E
and F are, of course, on an arc struck from the axis S, and it is
obvious that in whatever position the centre H may be it will be
somewhere on the arc I, I, which is struck from the centre S'. Now
suppose that D moves to E, and if we take the radius D, H, and from E
mark it upon the arc I as at V, then H will obviously be the new
position of H. To find the new position of G we first strike the arc J,
J, because in every position of G it will be somewhere on the arc J, J.
To find where that will be when H is at V, take the radius H, G, and
from V as a centre mark it on J, J, as at K, which is the position of G
when D is at E and H is at V. For the positions when D is at F we repeat
the process, taking the radius D, H, and from F marking P, and with the
radius H, G, and from P as a centre marking Q; then P is the new
position for H, and Q is that for G.
[Illustration: Fig. 263.]
In Figure 264 a lever arm A and cam C are in one piece on a shaft. S is
a shoe sliding on the line _x_, and held against the cam face by the rod
R; it is required to find the position of the face of the shoe against
the cam when the end of the arm is at D.
Draw line E from D to the axis of the shaft and line F. From the shaft
axis as a centre draw circle W; draw line J parallel to _x_. Take the
radius G H, and from K as a centre mark point P on W; draw line Q from
the shaft axis through P, and mark point T. From the shaft axis as a
centre draw from T an arc, cutting J at V, and V is the point where the
face of the shoe and the face of the cam will touch when the arm stands
at D.
[Illustration: Fig. 264.]
Let it be required to find the amount of motion imparted in a straight
line to a rod attached to an eccentric strap, and the following
construction may be used. In Figure 265 let A represent the centre of
the shaft, and, therefore, the axis about which the eccentric revolves.
Let B represent the centre of the eccentric, and let it be required to
find in what position on the line of motion _x_, the centre C of the rod
eye will be when the centre B of the eccentric has moved to E. Now since
A is the axis, the centre B of the eccentric must rotate about it as
denoted by the circle D, and all that is necessary to find the position
of C for any position of eccentric is to mark the position of B on
circle D, as at E, and from that position, as from E, as a centre, and
with the length of the rod as a radius, mark the new position of C on
the line _x_ of its motion. With the centre of the eccentric at B, the
line Q, representing the faces of the straps, will stand at a right
angle to the line of motion, and the length of the rod is from B to C;
when the eccentric centre moves to E, the centre line of the rod will be
moved to position P, the line Q will have assumed position R, and point
C will have moved from its position in the drawing to G on line _x_. If
the eccentric centre be supposed to move on to F, the point C will move
to H, the radii B C, E G, and F H all being equal in length. Now when
the eccentric centre is at E it will have moved one-quarter of a
revolution, and yet the point C will only have moved to G, which is not
central between C and H, as is denoted by the dotted half circle I.
[Illustration: Fig. 265.]
On the other hand, while the eccentric centre is moving from E to F,
which is but one-quarter of a revolution, the rod end will move from G
to H. This occurs because the rod not only moves _endwise_, but the end
connected to the eccentric strap moves towards and away from the line
_x_. This is shown in the figure, the rod centre line being marked in
full line from B to _x_. And when B has moved to E, the rod centre line
is marked by dotted line E, so that it has moved away from the line of
motion B _x_. In Figure 266 the eccentric centre is shown to stand at an
angle of 45 degrees from line _q_, which is at a right angle to the line
of motion _x x_, and the position of the rod end is shown at C, J and H
representing the extremes of motion, and G the centre of the motion.
[Illustration: Fig. 266.]
If now we suppose the eccentric centre to stand at T, which is also an
angle of 45 degrees to _q_, then the rod end will stand at K, which is
further away from G than C is; hence we find that on account of the
movement of the rod out of the straight end motion, the motion of the
rod end becomes irregular in proportion to that of the eccentric, whose
action in moving the eye C of the rod in a straight line is increased
(by the rod) while it is moving through the half rotation denoted by V
in figure, and diminished during the other half rotation.
In many cases, as, for example, on the river steamboats in the Western
and Southern States, cams are employed instead of eccentrics, and the
principles involved in drawing or marking out such cams are given in the
following remarks, which contain the substance of a paper read by Lewis
Johnson before the American Society of Mechanical Engineers. In Figure
267 is a side view of a pair of cams; one, C, being a full stroke cam
for operating the valve that admits steam to the engine cylinder; and
the other, D, being a cam to cut off the steam supply at the required
point in the engine stroke. The positions of these cams with relation to
the position of the crank-pin need not be commented upon here, more than
to remark that obviously the cam C must operate to open the steam inlet
valve in advance of cam D, which operates to close it and cause the
steam to act expansively in the cylinder, and that the angle of the
throw line of the cut-off valve D to the other cam or to the crank-pin
varies according as it is required to cut off the steam either earlier
or later in the stroke.
[Illustration: Fig. 267.]
The cam yoke is composed of two halves, Y and Y', bolted together by
bolts B, which have a collar at one end and two nuts at the other end,
the inner nuts N N enabling the letting together of the two halves
of the yoke to take up the wear. It is obvious that as the shaft
revolves and carries the cam with it, it will, by reason of its shape,
move the yoke back and forth; thus, in the position of the parts shown
in Figure 267, the direction of rotation being denoted by the arrow, cam
C will, as it rotates, move the yoke to the left, and this motion will
occur from the time corner _a_ of the cam meets the face of Y' until
corner _b_ has passed the centre line _d_. Now since that part of the
circumference lying between points _a_ and _b_ of the cam is an arc of a
circle, of which the axis of the shaft is the centre, the yoke will
remain at rest until such time as _b_ has passed line _d_ and corner _a_
meets the jaw Y of the yoke; hence the period of rest is determined by
the amount of circumference that is made concentric to the shaft; or, in
other words, is determined by the distance between _a_ and _b_.
The object of using a cam instead of an eccentric is to enable the
opening of the valves abruptly at the beginning of the piston stroke,
maintaining a uniform steam-port opening during nearly the entire length
of stroke, and as abruptly closing the valves at the termination of the
stroke.
Figure 268 is a top view of the mechanism in Figure 267; and Figure 269
shows an end view of the yoke. At B, in Figure 268, is shown a guide
through which the yoke-stem passes so as to be guided to move in a
straight line, there being a guide of this kind on each side of the
yoke.
[Illustration: Fig. 268.]
The two cams are bolted to a collar that is secured to the crank-shaft,
and are made in halves, as shown in the figures and also in Figures 270
and 271, which represent cams removed from the other mechanism. To
enable a certain amount of adjustment of the cams upon the collar, the
bolts which hold them to the collar fit closely in the holes in the
collar, but the cams are provided with oblong bolt holes as shown, so
that the position of either cam, either with relation to the other cam
or with relation to the crank-pin, can be adjusted to the extent
permitted by the length of the oblong holes.
[Illustration: Fig. 269.]
The crank is assumed in the figures to be on its dead centre nearest to
the engine cylinder, and to revolve in the direction of the arrows. The
cams are so arranged that their plain unflanged surfaces bolt against
the collar.
The method of drawing or marking out a full stroke cam, such as C in
Figure 267, is illustrated in Figure 272, in which the dimensions are
assumed to be as follows:
Diameter of crank shaft, 7-1/2 inches; travel of cam, 3 inches; width of
yoke, 18 inches.
[Illustration: Fig. 270.]
The circumference of the cam is composed of four curved lines, P, P', K
1, and K 2. The position of the centre of the crank shaft in this
irregularly curved body is at X. The arcs K 1 and K 2 differ in radius,
but are drawn from the same point, X, and hence are concentric with the
crank shaft.
The arcs P, P', are of like radius, but are drawn from the opposite
points S, S', shown at the intersection of the arcs P, P', with the
arc K 1. Thus arcs P, P', are eccentric to the crank shaft.
[Illustration: Fig. 271.]
[Illustration: Fig. 272.]
[Illustration: Fig. 273.]
To draw the cam place one point of the dividers at X, which is the
centre of the crank shaft, and draw the circle E equal to width of yoke,
18 inches. Through this centre X, draw the two right lines A and B. On
the line B, at the intersection of the curved line E, draw the two
vertical lines A 1, A 1. With a radius of 10-1/2 inches, and with one
point of the dividers at X, draw the arc K 1. With a radius of 7-1/2
inches, and one point of the dividers at X, draw the arc K 2. With a
radius of 18 inches, and one point of the dividers at the intersection
of the arc E, with the vertical line A 1 at S, draw the arc P opposite
to S, and let it merge or lose itself in the curved line K 2. Draw the
other curved line P' from the other point S, and we have a full stroke
cam of the dimensions required, and which is represented in Figure 273,
removed from the lines used in constructing it.
[Illustration: Fig. 274.]
The engravings from and including Figure 274 illustrate the lines
embracing cut-off cams of varying limits of cut-off, but all of like
travel and dimensions, which are the same as those given for the full
stroke cam in Figure 272.
In drawing cut-off cams, the stroke of the engine plays a part in
determining their conformation, and in the examples shown this is
assumed to be 4 feet. Figure 274 illustrates the manner of finding
essential points in drawing or marking out cut-off cams. With X as a
centre, and a radius of 2 feet, draw the circle E 1, showing the path of
the crank-pin in making a revolution. This circle has a diameter of 4
feet, equal to the stroke of the engine. Draw the horizontal line B,
passing through the centre of circle E 1. Within the limits of circle E
1, subdivide line B into eight equal parts, as at 1, 2, 3, 4, etc. Draw
the vertical lines, 1, 2, 3, 4, etc., until they each intersect the
circle E 1.
[Illustration: Fig. 275.]
With X as a centre, draw the circle E, having a diameter of 18 inches,
equal to the space in the yoke embracing the cam.
From the centre X draw the series of radial lines through the points of
intersection of the vertical lines 1, 2, 3, 4, etc., from the circle E
1, and terminating at X. We will now proceed to utilize the scale
afforded by Figure 274, in laying off the cut-off cam shown in Figure
276, of half stroke limit.
[Illustration: Fig. 276.]
With X as a centre, draw the circle E, Figure 275, having a diameter of
18 inches. Bisect this circle with the straight lines A and B, which
bear the same relation to their enclosing circle that the lines A, B,
do to the circle E in Figure 274.
[Illustration: Fig. 277.]
It will be observed, in Figure 274, that the vertical line A is (at the
top half) also No. 4, representing 4/8, or half of the stroke. With a
radius of 18 inches, and one point of the dividers placed at V, which is
at the intersection of the circle E with the horizontal line B in Figure
275, draw the arc P. With the same radius and with one compass point
rested at V', draw the arc P'; then two arcs, P and P', intersecting at
the point S.
With the same radius and one point of the compasses at S, draw the arc H
H. The arcs K 1 and K 2 are drawn from the centre X, with a radius of
10-1/2 for K 1 and 7-1/2 inches for K 2, and only serve in a half
stroke cam to intersect the curved lines already drawn, as shown in
Figure 275. In practice, the sharp corner at S would be objectionable,
owing to rapid wear at this point; and hence a modification of the
dimensions for this half stroke cam would be required to obtain a larger
wearing surface at the point S, but the cam of this limit (1/2 stroke)
is correctly drawn by the process described with reference to Figure
275, the outline of the cam so constructed being shown in Figure 276.
[Illustration: Fig. 278.]
In Figure 278 is shown a cam designed to cut off the steam at
five-eighths of the piston stroke, the construction lines being given
in Figure 277, for which draw circle E and straight lines A and B, as in
the preceding example. By reference to Figure 274 it will be observed
that the diagonal line drawn through circle E at 5 is drawn from the
straight line marked 5, which intersects circle E 1, and as this
straight line 5 represents five-eighths of the stroke laid off on line
B, it determines the limit of cut-off on the five-eighths cam in Figure
277.
[Illustration: Fig. 279.]
Turning then to Figure 274, take on circle E the radius from radial line
4 to radial line 5, and mark it in Figure 277 from the vertical line
producing V'.
Now, with a radius of 18 inches, and one point of the dividers fixed at
point V, forming the intersection of the circle E with the horizontal
line B, draw the arc P. With the same radius, and one point of the
dividers fixed at point V', draw the opposite arc P'. With a radius of
10-1/2 inches from the centre X, draw the arc K 1, intersecting lines P
P', at S S. With a radius of 7-1/2 inches, draw the curved line K 2,
opposite to curved line K 1. Now, with a radius of 18 inches, and one
point of the dividers fixed alternately at S S, draw the arcs H, H, from
their intersection with the circle E, until they merge into the curved
line K 2. These curved lines embrace a cut-off cam of five-eighths
limit, shown complete in Figure 278.
[Illustration: Fig. 280.]
From the instructions already given it should be easy to understand that
the three-fourths and seven-eighths cams, shown in Figures 279, 280, 281
and 282, are drawn by taking the points of their cut-off from the same
scale shown in Figure 274, at the diagonal points 6 and 7, intersecting
circle E in that figure; and cut-off cams of intermediate limit of
cut-off can be drawn by further subdividing the stroke line B, in Figure
274, into the required limits.
[Illustration: Fig. 281.]
Cut-off cams of any limit are necessarily imperfect in their operations
as to uniformity of cut-off from opposite ends of the slides, not from
any defect in the rule for laying them off, but from the well-known fact
of the crank pin travelling a greater distance, while driven by the
piston from the centre of the cylinder, through its curved path from the
cylinder, over its centre, and back to the centre of the cylinder, than
in accomplishing the remaining distance of its path in making a complete
revolution; and, although the subdivisions of eighths of the stroke
line B, in Figure 274, does not truly represent a like division of the
piston stroke, owing to deviation, caused by inclination of the
connecting rod in traversing from the centres to half stroke, still it
will be found that laying off a cut-off cam by this rule is more nearly
correct than if the divisions on stroke line B were made to correspond
exactly with a subdivision of piston stroke into eighths.
[Illustration: Fig. 282.]
The cut-off in cams laid off by the rules herein described is greater in
travelling from one side of the slides than in travelling from the
opposite end, one cut-off being more than the actual cut-off of piston
stroke, and the other less; and in practical use, owing to play or lost
motion in the connections from cam to valve, the actual cut-off is less
than the theoretical; hence cut-off cams are usually laid off to
compensate for lost motion; that is, laid off with more limit; for
instance, a five-eighths cam would be laid off to cut-off at
eleven-sixteenths instead of five-eighths.
[Illustration: Fig. 283.]
Figure 283 represents the motion a crank, C, imparts to a connecting
rod, represented by the thick line R, whose end, B, is supposed to be
guided to move in a straight line. The circle H represents the path of
the crank-pin, and dots 1, 2, 3, etc., are 24 different crank-pin
positions equidistant on the circle of crank-pin revolution. Suppose the
crank-pin to have moved to position 1, and with the compasses set to the
length of the rod R, we set one point on the centre of position 1, and
mark on the line of motion _m_ the line _a_, which will be the position
rod end B will have moved to. Suppose next that the crank-pin has moved
into position 2, and with the compass point on the centre of 2 we mark
line 2, showing that while the crank-pin moved from 1 to 2, the rod end
moved from _a_ to _b_; by continuing this process we are enabled to
discern the motion for the whole of the stroke. The backward stroke will
be the same, for corresponding crank-pin positions, for both strokes;
thus, when the rod end is at 7 the crank-pin may be at 7 or at 17. This
fact enables us to find the positions for the positions later than 6, on
the other side of the circle, as at 17, 16, 15, etc., which keeps the
engraving clear.
[Illustration: Fig. 284.]
In Figure 284 a pinion, P, drives a gear-wheel, D, on which there is a
pin driving the sliding die A in the link L, which is pivoted at C, and
connected at its upper end to a rod, R, which is connected to a bolt, B,
fast to a slide, S. It is required to find the motion of S, it moving in
a straight line, dotted circle H' representing the path of the pin in
the sliding die A, arc H representing the line of motion of the upper
end of link L, and lines N, O, its centre line at the extreme ends of
its vibrating motion. In Figure 285 the letters of reference refer to
the same parts as those in Figure 284. We divide the circle H' of pin
motion into 24 equidistant parts marked by dots, and through these we
draw lines radiating from centre, C, and cutting arc H, obtaining on the
arc H the various positions for end Z of rod R, these positions being
marked respectively 1, 2, 3, 4, etc., up to 24. With a pair of
compasses set to the length of rod R from 1 on H, as a centre, we mark
on the line of motion of the slide, line _a_, which shows where the
other end of rod R will be (or in other words, it shows the position of
bolt B in Figure 284), when the centre of A, Figure 284, is in position
1, Figure 285.
[Illustration: Fig. 285.]
From 2 on arc H, we mark with the compasses line _b_ on line M, showing
that while the pin moved from 1 to 2, the rod R would move slide S,
Figure 284, from _a_ to _b_, in Figure 285. From 3 we mark _c_, and so
on, all these marks being above the horizontal line M, representing the
line of motion, and being for the forward stroke. For the backward
stroke we draw the dotted line from position 17 up to arc H, and with
the compasses at 17 mark a line beneath the line M of motion, pursuing
the same course for all the other pin motions, as 18, 19, etc., until
the pin arrives again at position 24, and the link at O, and has made a
full revolution, and we shall have the motion of the forward stroke
above and that of the backward one below the line of motion of the
slide, and may compare the two.
[Illustration: Fig. 286.]
[Illustration: Fig. 287.]
[Illustration: Fig. 288.]
[Illustration: Fig. 289.]
Figures 286 and 287 represent the Whitworth quick return motion that is
employed in many machines. F represents a frame supporting a fixed
journal, B, on which revolves a gear-wheel, G, operated by a pinion, P.
At A is an arm having journal bearing in B at C. This arm is driven by a
pin, D, fast in the gear, G; hence as the gear revolves, pin D moves A
around on C as a centre of motion. A is provided with a slot carrying a
pin, X, on which is pivoted the rod, R. The motion of end N of the rod R
being in a straight line, M, it is required to find the positions of N
during twenty-four periods in one revolution of G. In Figure 288 let H'
represent the path of motion of the driving pin D, about the centre of
B, and H the path of motion of X about the centre C; these two centres
corresponding to the centres of B and C respectively, in Figure 287. Let
the line M correspond to the line of motion M in Figure 286. Now since
it is the pin D, Figure 287, that drives, and since its speed of
revolution is uniform, we divide its circle of motion H' into
twenty-four equal divisions, and by drawing lines radiating from centre
C, and passing through the lines of division on H' we get on circle H
twenty-four positions for the pin X in Figure 286. Then setting the
compasses to the length of the rod (R, Figure 286), we mark from
position 1 on circle H as a centre line, _a_; from position 2 on H we
mark line _b_, and so on for the whole twenty-four positions on circle
H, obtaining from _a_ to _n_ for the forward, and from _n_ to _y_ for
the motion during the backward stroke. Suppose now that the mechanism
remaining precisely the same as before, the line M of motion be in a
line with the centres C, B, instead of at a right angle to it, as it is
in Figure 286, and the motion under this new condition will be as in
Figure 289; the process for finding the amount of motion along M from
the motion around H being precisely as before.
[Illustration: Fig. 290.]
In Figure 290 is shown a cutter-head for a wood moulding machine, and it
is required to find what shape the cutting edge of the cutter must be
to form a moulding such as is shown in the end view of the moulding in
the figure. Now the line A A being at a right angle to the line of
motion of the moulding as it is passed beneath the revolving cutter, or,
what is the same thing, at a right angle to the face of the table on
which the moulding is moved, it is obvious that the highest point C of
the moulding will be cut to shape by the point C of the cutter; and that
since the line of motion of the end of the cutter is the arc D, the
lowest part of the cutter action upon the moulding will be at point E.
It will also be obvious that as the cutter edge passes, at each point,
its length across the line A A, it forms the moulding to shape, while
all the cutting action that occurs on either side of that line is
serving simply to remove material. All that we have to consider,
therefore, is the action on line A A.
It may be observed also that the highest point C of the cutter edge must
not be less than 1/4 inch from the corner of the cutter head, which
gives room for the nut N (that holds the cutter to the head) to pass
over the top of the moulding in a 2-1/2 inch head. In proportion as the
heads are made larger, however, less clearance is necessary for the nut,
as is shown in Figure 291, the cutter edge extending to C, and therefore
nearly up to the corner of the head. Its path of motion at C is shown by
dotted arc B, which it will be observed amply clears the nut N. In
practice, however, point C is not in any size of cutter-head placed
nearer than 1/4 inch from corner X of the cutter-head.
To find the length of the cutter edge necessary to produce a given depth
of moulding, we may draw a circle _i_, Figure 292, equal in diameter to
the size of the cutter head to be used, and line A A. The highest point
of cutting edge being at _e_, and the lowest at g, then circles _d_ and
_f_ represent the line of motion of these two points; and if we mark the
cutter in, the necessary length of cutting edge on the cutter is
obviously from _a_ to _b_.
[Illustration: Fig. 291.]
[Illustration: Fig. 292.]
Now the necessary depth of cutter edge being found for any given
moulding, or part of a moulding, the curves for the edge may be found as
follows: Suppose the moulding is to be half round, as in the end view in
Figure 290. The width of the cutter must of course equal the width of
the moulding, and the length or depth of cutting edge required may be
found from the construction shown in Figure 292; hence all that remains
is to find the curve for the cutting edge. In Figure 293, let A A
represent the centre of the cutter width, its sides being F F', and its
end B B. From centre C draw circle D, the upper half of which will serve
to represent the moulding. Mark on A the length or depth the cutting
edge requires to be, ascertaining the same from the construction shown
in Figure 292, and mark it as from C to K'. Then draw line E E, passing
through point K. Draw line G, standing at the same angle to A A as the
face _h b_, Figure 292, of the cutter does to the line A A, and draw
line H H, parallel to G. From any point on G, as at I, with radius J,
draw a quarter of a circle, as K. Mark off this quarter circle into
equal points of division, as by 1, 2, 3, etc., and from these points of
division draw lines, as _a_, _b_, _c_, etc.; and from these lines draw
horizontal lines _d_, _e_, _f_, etc. Now divide the lower half of circle
D into twice as many equal divisions as quarter circle K is divided
into, and from these points of division draw perpendiculars _g_, _h_,
_i_, etc. And where these perpendiculars cross the horizontal lines, as
_d_, will be points through which the curve may be drawn, three of such
points being marked by dots at _p_, _q_, _r_. If the student will, after
having drawn the curve by this construction, draw it by the construction
that was explained in connection with Figure 79, he will find the two
methods give so nearly identical curves, that the latter and more simple
method may be used without sensible error.
[Illustration: Fig. 293.]
[Illustration: Fig. 294.]
When the curves of the moulding are not arcs of circles they may be
marked as follows:
Take the drawing of the moulding and divide each member or step of it by
equidistant lines, as _a_, _b_, _c_, _d_, _e_, _f_, _g_, in Figure 294;
above the moulding draw lines representing the cutter, and having found
the depth of cutting edge for each member by the construction shown in
Figure 292, finding a separate line, _a b_, for each member of the
moulding, transfer the depths so found to the face of the cutter; divide
the depth of each member of the cutter into as many equal divisions as
the corresponding member of the moulding is divided into, as by lines
_h_, _i_, _j_, _k_, _l_, _m_, _n_. Then draw vertical lines, as _o_,
_p_, _q_, _r_, etc.; and where these lines meet the respective lines
_h_, _i_, _j_, etc., are points in the curve, such points being marked
on the cutter by dots.