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CLOCK AND WATCH ESCAPEMENT MECHANICS (1997) - page 2

33

34

In calculus, the total work done is found by integration, that is, by finding the area

under the curve. I approximated the area by using Simpson's Rule. The Brocot graph

divided the time index into 50 parts, accurate to 2 decimal places. By dividing it into

more parts, I could get a more accurate approximation: dividing it into 10,000 parts, easy

to do on a computer, gave a work done index of 0.3927. If you would like to try this on

your computer, use the formulas on the page after the Brocot graph.

A practical example would demonstrate the usefulness of this information. A friend

asked me to look at the escapement of a Seth Thomas regulator wall clock with

Graham-style pallets. By inspection, I could see that the design was based on a square, so

the impulse face's angle of the entry pallet should have been horizontal and the impulse

face's angle of the exit pallet should have been vertical. On the exit side, the impulse face

was almost vertical, so I decided to leave it alone. On the entry side, the impulse face was

almost parallel to Fe. I estimated the angle FeFi to be about 70º.

Fi = 1 x cos (70) = 0.342

Fp = 0.342 x cos (20) = 0.321

This does not account for drop, so compare this efficiency of 32% with 50%. The clock

was less efficient by more than a third on the entry side because the impulse face's angle

was incorrect. Since there was plenty of lock, I decided to change the impulse face's

angle by as much as I could without sacrificing all the lock. I did this using an Arkansas

stone, installing the pallet frequently to check the lock. The result was not horizontal, but

an improvement over the previous angle. Then I polished, cleaned and oiled the impulse

face. By decreasing the angle FeFi by about 10º, there was an improvement in efficiency:

Fi

= cos (60) = 0.5

Fp = 0.5 cos (30) = 0.433

Efficiency improved by over 20%. This example demonstrates how this information is

useful to the repairman. While it may not be possible to adjust the impulse face's angle

by as much as would be ideal, an improvement is surely better than none.

The most difficult part of an escapement

drawing is the calculations for the verge cir-

cle radii, which is critical for the Pin Wheel

escapement. Here there are seven circles (not

drawn to scale). If the 4th circle, between the

pallets, has a radius of 6", the others must be

1

7

calculated from this circle. If the escape

wheel's pin has a diameter of 1/16 of an inch,

or 0.0625", the gap between the two pallets

must be greater than this, say 0.1": the radii

of the 3rd and the 5th circles would be 5.95"

and 6.05" respectively. The gap between the

locking face of the entry pallet and the let-off

corner of the exit pallet must be equal to the

arc between two escape wheel pins (less the diameter of one pin and less the allowance

for drop). If the escape wheel had fifteen pins and a radius of 3", its arc would be:

arc = 3 x (24º x pi) / 180 = 1.26"

(because 24º x 15 = 360º), but you allow 1.5º for drop:

arc = 3 x (22.5 x pi) / 180 = 1.18"

and for the pin diameter: arc = 1.18 - 0.0625 = 1.12" (to two decimal places).

The radius of the 1st circle is therefore:

6 - (1.12 / 2) = 5.44"

35

and the radius of the 7th circle is 6.56".

The radius of the 2nd circle is half way between the 1st and the 3rd:

(5.95 + 5.44) / 2 = 5.70"

and the radius of the 6th circle is half way between the 5th and the 7th:

(6.05 + 6.56) / 2 = 6.30"

The radius of the 4th circle may appear to be arbitrary because it is not determined

by the tooth span between the pallets, as it would be for the Graham, Recoil, and Brocot.

It is determined only by the desired angle of pendulum swing.

Consider a pallet circle that was determined by a tooth span, such as the Graham

escapement with the 15 tooth escape wheel in chapter 10. It had a tooth span of 5.5 teeth,

so the angle between the pallets was: (5.5 / 15) x 360 = 132º, half of which was 66º. The

escape circle radius was 3", so the pallet circle radius could be found by triangulation:

tan (66) = Rp / Re

Rp = Re x tan (66) = 3 tan (66) = 6.74"

Rp = ?

Continuing with the Graham in chapter 10, its escape wheel had

a radius of 3", so the arc between two teeth was:

arc = (3" x 24º x pi) / 180º = 1.26"

90º

The arc per beat (12º): 1.26 / 2 = 0.63"

Allow 1.5º for drop:

arc = (0.63 x 10.5º) / 12º = 0.55"

This is the thickness of the pallets.

66º

Re = 3"

Therefore, the entry pallet locking face has a pallet circle radius of:

6.74 + 0.27 = 7.01"

and the exit pallet locking face has a pallet circle radius of :

6.74 - 0.27 = 6.47".

Replacing the Pallets.

Suppose a British grandfather clock

movement were brought for repair with no

pallets. You determine by inspecting the 30

tooth escape wheel that the clock had a recoil

escapement. The distance between the escape

wheel bushing and the pallet bushing is 1.6".

The escape wheel has a diameter of 2", so the

escape circle's radius is 1". First, draw the pal-

let you want over an ideal Graham pallet and

remove the latter:

36

Draw two triangles using this arrangement:

cos (A) = 1 / 1.6

A = acos (1 / 1.6) = 51.3º

1.6

90

Use this to determine the tooth span:

30 x (2 x 51.3 / 360) = 8.553

1

A

so you should use a span of 8.5 teeth.

To recalculate the angle A: (8.5 x 360) / (2 x 30) = A = 51º

1.59

90

1

51

Assume a small error in measurement:

39

39

cos (51) = 1 / h

1.23

1.23

h = 1 / cos (51) = 1.59"

90

1.59

90

The pallet angle is : 180 - 90 - 51 = 39º

The pallet circle radius is:

1

1

Rp = 1 x tan (51) = 1.23

51

51

By comparing this drawing with the one on the

previous page, calculate the positions of the tips of

the pallets. The inside tip of the entry pallet needs to

be at a slightly smaller angle from vertical:

Angle = 39 - (3 x (1 / 1.23) = 36.6º

and the radius:

r = 1.23 - ((2 x pi) / (4 x 30)) = 1.18"

39

39

1.23

1.23

Similarly for the exit pallet:

Angle = 36.6º.

90

90

1.59

r = 1.23 + ((2 x pi) / (4 x 30)) = 1.28"

The rest of the pallet could be drawn any way

1

1

51

51

you wish. To explain the numbers above, the escape

wheel has an angle of 12º between two teeth. A tooth

will travel 6º per beat. The pallet angle lines were di-

vided into 3º on either side. Since the pallet radius is

1.23 and the escape radius is 1", multiply the escape

angle of 3º by the ratio of the radii. Since the pallet

radius is larger, the pallet arc is smaller than the es-

cape wheel arc (which has an angle of 3º).

37

For the radius calculation, decrease the

pallet radius on the entry pallet by the

amount of a quarter of the arc between two

escape wheel teeth, and increase the pallet

radius on the exit pallet similarly. This is

only a close approximation because the

change in the pallet circle's radius is as-

sumed to be equal to the arc of the escape

36.6

36.6

1.18

1.28

circle, but an arc is not a straight line.

Once the pallet blank is made, it must

be fitted because there is no drop. By filing

1.59

on the surfaces pointed to by the arrows,

some inside and outside drop could be cre-

ated.

When designing a Graham pallet, the issue becomes more complicated because of

lock. I would suggest that the best way to design the pallets would be to make the blank

with impulse faces that have angles relative to Fe smaller than 45º, such as 25º. When

fitting the pallets to the clock, file first on the inside and outside surfaces to create

enough inside and outside drop. Then file on the impulse face to increase the impulse

angle until the lock has been reduced to about 1º. To be practical in a manufacturing

situation, the measurement deviations must be accounted for to keep rejects statistically

below 5%. The need to account for these deviations means that a compromise situation

arises because fitting each verge by hand would interfere with mass production and

economies of scale.

Fitting a Brocot pallet requires the calculation of the pallet circle's radius. If the

escape wheel rotates 6º per beat, and its radius were 1", the arc would be:

arc = (1 x 6 x pi) / 180 = 0.105"

I suggest 1.5º of drop, so the radius of the pallet's circle should be:

r = (1 x 4.5 x pi) / 180 = 0.08"

Using the recoil example, the Brocot pallet should be positioned such that the point

where the pallet releases the escape tooth should have a pallet circle radius of 1.18" and

an angle of 36.6º from vertical for the entry pallet, and a radius of 1.28" and an angle of

36.6º for the exit pallet. Then the lock and drop could be further adjusted by raising or

lowering the pallets, or by changing the gap between the pallets.

By studying the above example, you could see how you could use one type of

escapement for the calculations, and then apply them to the type of escapement you want

to create. A step-by-step process makes the calculations easier.

38

3º

where the vertical

(Fig. 1)

line

(1) intersects

(4)

with line (4), rotate

(5)

line

(4) clockwise

by 3º: this will de-

termine the amount

of lock that this

design will have.

While the ideal

45º

45º

amount of lock is

(3)

recommended to

24º

be only 1º, I rec-

(2)

(1)

ommend 3º in this

24º

drawing because it

(6)

24º

30º

will make the lock

(7)

easier to see during

4"

6"

7.5"

the simulation. It

diameter

(8)

will also make the

simulation

more

forgiving if errors

were made during

the preparation of

the drawing. The

best way to ensure

that line (4) is ro-

tated about the

point where it in-

tersects line

(1) is

to center line

(4)

on the page shown on your computer screen before rotating, since line (1) has already

been centered on the page.

Rotate line (2) clockwise by 24º to get line (6), and place it onto the point where lines

(2), (3), and (4) intersect. Line (6) will be the tooth's locking face, and the point where it

intersects the other three lines will be the entrance corner of the tooth.

Rotate line (2) counterclockwise by 24º, since there are 24º between each tooth of a

15 tooth escape wheel, to get line (7). Rotate line (7) clockwise by 24º to get line (8).

Place line (8) onto the point where line (7) intersects the six inch circle. Line (8) will be

the locking face of the next tooth. The locking faces of the teeth will appear to lean

forwards by 24º.

Draw a three inch circle and place in onto the point where lines (3) and (5) intersect.

This circle can be made larger or smaller, or stretched at will until the desired shape is

achieved to trace the curve of the tooth. I recommend that you draw a relatively thin tooth

that would allow the pallet to clear during the simulation. After the simulation, when you

know how much clearance you have, you could draw the escape wheel again.

42

Draw the first tooth over lines (3) and (8), and trace the

curve over the ellipse, so that you get a tooth that looks like this:

Move all the lines and the two inner circles to

one side, so that all you have on the page is the

7.5 inch circle and the tooth in the same position.

Group the tooth and the circle, and rotate the

group by 24º in whichever direction you choose.

Duplicate the group and rotate again. Duplicate the

group again and rotate again. Repeat this until you

have 15 groups, all in the same place. Then ungroup

all 15 groups, so that you end up with 15 teeth and

15 circles. Delete 14 of the circles, one by one, until

you have the complete wheel.

In order for the escape wheel to be true

(or

perfectly centered), it is necessary to rotate the tooth

inside a circle that extends beyond the tooth. Since

the tooth extends slightly beyond the edge of the six

inch circle, a larger circle is needed for the drawing.

43

13: Drawing the Pallets.

(4)

90º

The first tooth was drawn next to line

(1)

(2), which had an angle of 30º relative to

(2)

30º

the vertical line

(1). The

30º angle was

chosen so that there would be a 2.5 tooth

span between the pallets for a

15 tooth

wheel. The angle between lines (2) and (4)

was 90º so that the angle between the force

exerted by the escape tooth and the force

that acted to rotate the pallet would be 90º.

(10)

(4)

pallet circle

center

90º

90º

(1)

Rotate line (1) clockwise by 30º to get

(2)

(9)

line (9). Rotate line (9) by 90º to get line

30º

30º

(10). Place line (10) onto the point where

line (9) intersects the edge of the six inch

circle. The point where lines (1), (4) and

(10) intersect will be the pallet's circle

center.

1º

If you rotate line (2) counterclockwise by 2º, you

1º

would see that the escape wheel's impulse face occupies

a span of 2º of escape wheel rotation. (The impulse

face also provides 3º of lift and lock.)

3º

44

Since the escape wheel rotates by 12º per beat and you want 2º for drop and 2º for

the escape tooth's impulse face, the pallet should occupy a span of 8º. Rotate line (9)

counterclockwise by 4º to get line (11). Rotate line (9) clockwise by 4º to get line (12).

Repeat with line (2) to get lines (13) and (14):

(10)

(4)

4º

4º

4º

4º

(2)

(9)

(14)

(11)

(13)

(12)

45

Rotate line (2) counterclockwise by 45º to get line (15). Notice that it is at an angle

half way between the escape circle's radius and the pallet circle's radius, shown by lines (2)

and (4) respectively:

(10)

(4)

(15)

45º

45º

(2)

(9)

30º

30º

46

Rotate line (13) clockwise by 15º to get line (16),

(17)

and place it onto the point where lines (13) and (15)

(18)

intersect. Line (16) will be the pallet's locking face, and

90º

it will have a draw angle of 15º. Duplicate line (16) and

place it onto the point where lines

(14) and

(15)

intersect: lines (16) and (17) are parallel. Rotate line

(17) by 90º to get line (18) and place it in a suitable

position along lines (16) and (17). The pallet is now

(15)

recognizable. You could place a small line at the point

where lines (2) and (15) intersect: this small line would

show the mid-point of the pallet.

15º

(16)

(13)

(10)

(4)

(16)

(17)

(18)

(15)

(2)

(9)

(14)

(11)

(13)

(12)

47

Once you finish the entry pallet, group the lines and duplicate them. Rotate the sec-

ond pallet clockwise by 60º because that is the angle between lines (2) and (9), and then

place the pallet on the exit side such that the mid-point of the impulse face lies on the

point where lines (9) and (10) intersect. Most watches have different pallets, but an es-

capement using the same pallet on both sides can be designed, and using the same pallets

makes manufacture easier.

It will be necessary to draw the pallet circle to perform a simulation. The circle must

be larger than the drawing of the pallets, so I chose a six inch diameter circle, placed a

horizontal and a vertical line in it, grouped the circle and the lines, and then placed them

such that the circle center would lie on the point where lines (1), (4) and (10) intersect.

(10)

(4)

(1)

(2)

(9)

(14)

(11)

(13)

(12)

48

To draw the pallet arms, you could use circles,

stretched and shaped to give the outline you desire. It

should be obvious by looking at the drawing below that

the design is of equidistant impulse.

(10)

(4)

(2)

(9)

(14)

(11)

(13)

(12)

49

14: Changing the Design.

A different approach is required to create different designs, such as the equidistant

lock drawing below. Rotate lines (14) and (12) clockwise by 4º to get lines (15) and (16)

respectively. Rotate lines (10) and (4) clockwise by 7º to get lines (18) and (20), and

counterclockwise to get lines (17) and (19). Draw the entry pallet's impulse face from the

point where lines (2) and (20) intersect to the point where lines (15) and (19) intersect.

Draw the exit pallet's impulse face from the point where lines (9) and (17) intersect to the

point where lines (16) and (18) intersect. Rotate lines (2) and (9) clockwise by 15º to

draw the locking faces of the pallets. In order to maintain the action of the pallets as sym-

metrical as possible (and to keep the impulse face's angle as close to 45º as possible), the

lift is reduced slightly to 14º and the impulse angles are no longer at 45º, which reduces

efficiency, but the lock and drop are the same on both sides and the pallets are not "out of

angle." The pallets are no longer identical.

7º

7º

(18)

(19)

7º

7º

(10)

(4)

(17)

(20)

(13)

(16)

(2)

(12)

(14)

(9)

(15)

(11)

4º

4º

4º

4º

4º

4º

50

Drawings could be superimposed for comparison, such as this comparison between

the equidistant impulse and lock designs. It reveals that, in the equidistant impulse design,

the exit pallet's impulse face does not exactly fit between the lines: while the exit pallet is

identical to the entry pallet, it is not a mirror image of the entry pallet. The entry pallet

does not have a perfect fit either, but the difference is very small and inconsequential. The

result of the unequal fit on both sides is that the exit pallet releases the escape tooth a little

prematurely, causing the design to be out of angle: the angle on the left side to the

drop-lock position is one degree greater than on the right side. This problem is corrected

in Chapter 22.

L

I

I

L

entry

exit

side

side

51

To combine the advantages of equidistant impulse and equidistant lock, watchmakers

have designed an escapement half way between the two, called the semi-tangental es-

capement. Here, the angle between lines (2) and (23) and between lines (9) and (21) is 2º,

so rotate lines (2) and (9) counterclockwise by 2º to get lines (23) and (21). Rotate lines

(12) and (14) clockwise by 2º to get lines (22) and (24). Draw the entry pallet's impulse

face from the point where lines (23) and (20) intersect to the point where lines (24) and

(19) intersect. Draw the exit pallet's impulse face from the point where lines (21) and (17)

intersect to the point where lines (22) and (18) intersect. Rotate lines (23) and (21)

clockwise by 15º to draw the locking faces of the pallets.

7º

7º

(18)

(19)

7º

7º

(10)

(4)

(17)

(20)

(23)

(22)

(2)

(12)

(14)

(9)

(24)

(21)

2º

2º

4º

2º

2º

4º

52

15: Improving the Design.

In order to maximize efficiency, the impulse face of each pallet should have an angle

half way between its corresponding escape circle radius line and pallet circle radius line.

Looking at the design with equidistant impulse, if the angle between lines (2) and (4) were

90º, then the impulse face's angle should be 45º and line (15) is half way between lines (2)

and (4). A change in design is needed in practice because the escape wheel's teeth are too

narrow and therefore too fragile. Thus far, I have assumed that the escape tooth's impulse

face should be parallel to the pallet's impulse face in the drawing. However, the simulation

reveals, as suspected, that the lines do not remain parallel during impulse:

This means that they need not be parallel in the design. If the pallet's impulse face re-

mained unchanged, the efficiency of the escapement would remain unchanged, and we

could change the shape of the escape tooth without sacrificing efficiency. The span of the

tooth's impulse face could be increased from 2º to 4º so as to draw a wider tooth. The

amount of lock would remain unchanged at 3º for the purpose of the drawing:

2º

2º

1º

1º

3º

3º

34º

45º

If the escape tooth occupied a span of 4º and there were 2º of drop, the pallet would

occupy a span of 6º, for a total span of 12º per beat. New pallets must be drawn.

53

Rotate line (9) counterclockwise by 3º to get line (25), and clockwise by 3º to get line

(26). Rotate line (2) counterclockwise by 3º to get line (27), and clockwise by 3º to get

line (28). Rotate line (27) clockwise by 15º and draw the locking face. Draw the rest of

the pallet as before, then duplicate it and rotate it clockwise by 60º, placing it on the exit

side. The pallets are narrower, but the teeth are wider. Notice the difference in the angles

of the pallet and escape tooth impulse faces.

(10)

(4)

(2)

(9)

(28)

(25)

(27)

(26)

54

Another way to make a stronger tooth would be to trace the path of the pallet during

the simulation, and then to draw the curve on the back side of the tooth again. If the

simulation were superimposed for the action of the escapement as the tooth pushed on the

entry pallet,

the exit pallet enters the space between two teeth. The distance

between the exit pallet and the tooth that just passed it could be

measured by drawing lines from the pallet to the tooth. These lines

could be used as a guide when drawing a new curve. Move the ellipse

to other positions and change its shape until you trace a path for a

new curve more like the path of the pallet. If you draw the lines on the

tooth next to the exit pallet and are drawing the new tooth next to the

entry pallet, you must rotate the lines by 24º for every tooth in

between, that is, by 72º counterclockwise. I drew a new escape wheel

using the same principles as in the first drawing plus these lines as a

guide. Unfortunately, the path of the pallet would be very difficult to

predict before doing a simulation.

It is preferable to draw the first escape wheel with narrow teeth because it would be

much more likely to work in a simulation. For the same reason, I recommend designing

the first drawing with no less than 2º of lock and 2º of drop. The simulation would show

the lock and drop more clearly and it would be more forgiving if errors were made. In

addition, the simulation would be much more likely to work when changes in design are

made, such as when going from equidistant impulse to equidistant lock. Once you have

some practice, then try a design with the theoretically correct 1º of lock and 1º of drop.

You would find that any error made in the drawing would result in binding or recoil.

55

This drawing has the same, smaller pallets and a new escape wheel:

(10)

(4)

(2)

(9)

(28)

(25)

(27)

(26)

This drawing does not include the pallet fork and the roller table, primarily because of

the math involved. These are added in Chapter 22.

56

16:The English Lever.

This escapement has several disadvantages. The escape wheel's teeth are pointed, so

they lack the strength of the club-tooth design. The escape wheel's teeth do not have

impulse faces of their own, so the lock must be created by changing the design of the

pallets. This results in the pallets having impulse faces with angles other than the optimal

angle, so the design is less efficient.

The escape tooth is simple. The locking face appears to lean forwards by 24º. There

is an angle of 12º between the front and the back of the tooth, and there is a small gap to

create some thickness.

As in the Swiss Lever, there is 30º between the vertical line (1) and the escape circle

radius lines, (2) and (3), which are used to design the pallets. Rotate line (2) by 90º to get

line (4). Rotate line (3) by 90º to get line (5). Place the pallet circle such that its center lies

on the point where lines (4) and (5) intersect.

(5)

(4)

pallet

circle center

90º

90º

(1)

(2)

(3)

30º

30º

57

5º

5º5º

5º

The 15 tooth escape wheel rotates by

12º per beat, so make the pallet occupy a

span of 10º, in order to allow 2º for drop.

To draw the impulse

face lines, it is necessary

to determine their posi-

8º

2º

tions. Rotate line (4) by

8º

2º

(for lock) counter-

2º

clockwise to get line

(10). Then rotate line

(10) by

8º counter-

clockwise

to get line

(11). Rotate line (5) by

8º counterclockwise to

(13)

get line (12), and by 2º

(4)

(12)

more to get line

(13).

(1)

Draw a line from the

(2)

(3)

(7)

(8)

(5)

(10)

(6)

(9)

point where lines (6) and

(11)

(10) intersect to where

lines (7) and (11) inter-

sect: this will be the entry

pallet's

impulse face.

Draw a second line from

the point where lines (8)

and

(12) intersect to

where lines (5) and (9)

intersect: this will be the

exit pallet's impulse face.

58

(15)

(14)

(16)

15º

(17)

15º

(8)

(9)

(7)

(3)

(6)

(2)

8º

2º

8º

2º

Rotate line

(6)

clockwise by 15º to get

line (14), which will be

the entry pallet's lock-

ing face with a draw

angle of 15º. Duplicate

line (14) and place it on

(13)

the point where lines

(4)

(12)

(1)

(7) and

(11) intersect.

(2)

(3)

(7)

(8)

(5)

Repeat this procedure

(10)

(6)

(9)

for the exit pallet: ro-

(11)

tate line

(8) clockwise

by 15º to get line (16).

Duplicate line (16) and

place it on the point

where lines (5) and (9)

intersect. Finish the

drawing as you wish.

59

17: The Pin-Pallet Escapement.

You will recognize this

drawing from chapter 9 of

the Clock section. In order

to avoid repetition, I will

discuss only how it differs

from the modern Swiss

Lever Escapement.

In the club-tooth de-

sign, the escape circle radius

of each tooth meets the im-

pulse face and the locking

face at the entrance corner.

In the pin-pallet wheel,

however, the escape circle

radius of each tooth bisects

the impulse face. The

method used for the

pin-pallet wheel is the

correct approach, and it

must be used here. I used

the other method for the

club-tooth escape wheel be-

cause it enabled me to place

the entry pallet in line with

the tooth's impulse face at

the mid-point of the im-

pulse: in other words, to

create a better-looking drawing. The tooth's impulse face was less important because the

relationship between the angles changed during impulse, (see chapter 3). In fact, the last

drawing of the Swiss Lever showed tooth and pallet impulse faces that were not parallel.

The important criteria in the Swiss Lever are the angles of the pallet's impulse and locking

faces, and that the pallet's impulse face always be bisected by the pallet circle's radius line.

In the pin-pallet escapement, the angle of impulse is determined by the tooth's impulse

face rather than the pallet's impulse face, so the tooth's impulse face should always be bi-

sected by the escape radius line, and the angle between these two lines should always be

45º. The pallets are bisected by the pallet circle's radii. The pallets are also bisected by the

escape circle's radii, in order to have equidistant impulse. The amount of lock is deter-

mined by the radius of the pallet pin.

Both escapements were designed with the same number of escape wheel teeth and

with the same number of teeth between the pallets. Therefore, the other design principles

apply to both designs.

60

18: The Cylinder Escapement.

Rotate line (1) counterclockwise by 5º to get line (2)

and clockwise by 5º to get line (3). The tooth will occupy a

12º

span of 10º, and the escape wheel will rotate 12º per beat,

so there will be (no less than) 2º left over for cylinder thick-

ness and for drop. Rotate line (1) counterclockwise by 60º

5º

5º

to get line (4); place it on the point where line (1) and the

(5)

six inch diameter circle intersect. Rotate line

(2)

counterclockwise by 12º to get line (5); place it on the

60º

point where lines (2) and (4) intersect. Line (4) will become

the tooth's impulse face. Line (5) will become the back side

of the tooth. Draw a curve between them to form a triangle,

(4)

but give the entrance corner a slightly rounded edge. The

curve could be drawn by tracing it over a six inch diameter

(2)

(3)

circle.

(1)

Rotate the tooth in the 7.5

inch circle by 24º. Draw a curve

to connect the two teeth by trac-

ing over a small circle. Then draw

the escape wheel by duplication

and rotation, as before.

The cylinder will be a circle

with two crossing lines and a

thick curve traced over the edge

of the circle.

The thick curve will cover just over half the cir-

cumference of the circle. The inside diameter of the

curve will be slightly greater than the diameter of the

tooth. I chose a circle diameter of 0.73 inches (by trial

(4)

and error). Place the cylinder such that its center lies on

the point where the six inch diameter circle and lines

(2)

(3)

(1) and (4) intersect.

(1)

61

(2)

(3)

(1)

4 inch

6

7.5

diameter

Increasing the diameter of the cylinder circle results in a proportional increase in the

inside drop and an equal decrease in the outside drop.

The tooth impulse face's angle and length determine the lift. The cylinder occupies

just over half a circle in order to create lock: if it occupied half a circle, there would be no

lock.

A tooth impulse face angle of 45º would be more efficient, but this tooth does not fit

inside the cylinder as well as a tooth designed with an impulse face angle of 30º because it

would result in unequal inside and outside drops. In practice, cylinder escapements appear

to have escape tooth impulse faces with angles of less than 30º.

62

19: The Duplex Escapement.

All the drawings thus far have not included the action of the balance wheel, in order

to simplify the drawings and the simulations. The inclusion thereof complicates everything,

but is certainly necessary in this drawing.

This escapement should be thought of as having

two escape wheels, hence its name. Draw the first es-

1 inch

6 inches

cape wheel in a six inch diameter circle, each tooth

diameter

having a perpendicular locking face and a height of

one inch. This escape wheel has 15 teeth, so there are

24º between each tooth.

The inner escape wheel has smaller teeth, but they are

similar and also have perpendicular locking faces. Draw it

0.33

4.95 inch

in a 4.95 inch diameter circle, each tooth having a height

diameter

of a third of an inch. Both escape wheels need to be

designed so that they could be rotated either together or

independently.

The difficult part is calculating the relationships of the variables: how far from the es-

cape circle center the locking jewel's circle center needs to be, and the diameter of the lat-

ter. If the locking span were taken to be 10º, and the locking jewel's circle center were

placed on a circle 3.25 inches away from the escape circle center, its diameter could be

calculated.

J

J

The radius of the locking jewel would be:

45º

45º

3 sin 5 = J sin 45

J = 3 sin 5 / sin 45 = 0.370"

if the locking jewel rotated

90º during lock.

Draw a circle with a radius of 0.37 inches, and

draw a jewel in it as you wish.

3"

3"

5º

5º

63

Since the angle between two

escape teeth is 24º, rotate one es-

cape wheel by 12º so that, when

combined, the inner teeth would

appear to be half way between

each pair of outer teeth. Place the

two escape wheels together inside

a larger circle with a radius of 3.25

inches. Place the locking jewel

such that its center lies on the cir-

cumference of the larger circle.

Notice that the escape wheel ro-

tates by 10º during lock.

5º

5º

The impulse arm's length could be found by trial and error, or it could be calculated.

In this example, the impulse arm rotates by 60º during impulse. The escape wheel rotates

by 24º, less 2º for drop, or 22º. The impulse arm's circle radius will be given by 'X', and the

inner escape wheel's circle radius by 'W'.

Y / X = sin 30 = 0.5

X

30º

30º X

Y / W = sin 11 = 0.191

Y

Y

so,

0.5 X = 0.191 W

X = 0.382 W

W

W

X cos 30 + W cos 11 = 3.25

11º

11º

(0.382 W)(0.866) + W(0.982) = 1.312 W = 3.25

W = 2.476

(and 2W = escape circle diameter = 4.952 inches)

X = 0.382 x 2.476 = 0.946

(and 2X = impulse arm's circle diameter = 1.892 inches)

64

Draw the impulse arm's circle with two lines to indicate its center and with a diameter

of 1.89 inches. Place it such that its center lies on the center of the locking jewel. Rotate

the outer escape wheel clockwise until a tooth touches the locking jewel. Then rotate the

inner escape wheel until the impulse arm's circle becomes centered between two inner es-

cape teeth. Group the two escape wheels so that they would rotate as one.

65

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