This manual came with a Amsler Integrator 627/499, sold by Keuffel & Esser under number 3078-S.
Apart from the manual, there were a supplement and two typewritten notes.      # integrators

## type 626

 With 3 measuring units suitable for evaluating 1. the Area ∫ y dx = A 2. the Static Moment 1⁄2 ∫ y2 dx = M 3. the Moment of Inertia 1⁄3 ∫ y3 dx = I

### Placing the Integrator

Place the rail upon the drawing to be evaluated. Then arrange the Integrator in such a way that the two wheels run in the groove of the rail, and the integrating discs rest on the paper. The two gauges which are supplied with the instrument are then set in such a way that their edges hold in the groove of the rail and their points rest on the drawing. Now move the rail backwards or forwards until the points of the gauges rest exactly on the axis of moments x – x (i.e., on that line on which are to be based the static moments and the moments of inertia). The rail is then at the proper distance and parallel to the axis of moments. To obtain this position accurately if is advisable to use a magnifying glass.

The drum of each integrating disc is divided into 100 parts. Tenths of a part are read on the vernier. Complete revolutions are shown on the decadic drum which advances one line at every such turn. The decadic drum is arranged either for 10 or for 50 revolutions of the integrating disc.

Each complete reading is a figure either of four or of five digits, the thousands being read on the decadic drum the hundreds and tens on the drum, and the units on the vernier. Fig. 1

The reading in the adjoining figure, for exemple is 6569. The division of the vernier to be read, i.e., in the present example the pine is the division which is just opposite, or nearest, to a division of the roller. When taking the second reading after the first measurement, be careful to ascertain whether the motion of the integrating disc has been forward or backward, and how many times, and in what revolving direction the zero of the decimal drum has passed the fixed index mark, no account being taken of the short to and from passages through the index mark. If the total travel of the integrating disc has caused more than one complete turn of the decadic drum, the figure 10000 must be added to the difference of readings, as often as the counting disc has gone round. If the motion of the disc has been backward, a corresponding multiple of 10000 must be added to the initial reading before taking the difference between the initial and final readings.

### Circumscribing a figure

Make a mark on the outline of the figure to be measured. Set the fixed tracing point on the mark and write down the readings. Move the tracing point carefully along the outline of the figure in the clockwise direction till it comes back to the starting position. Take the second readings and write them down below the corresponding first ones. Subtract the first from the second readings and write down the difference to the right of the corresponding readings.

The figures then express the travel of the respective integrating discs.

### Tracing points

In addition to the fixed tracing point, the tracing arm of the Amsler Integrator Type 626 is provided with a moveable (vertically sliding) point No. 2, which is of value for measuring small figures.

Wherever practicable, trace diagrams of small height with the moveable points No. 2, in order to obtain a greater travel of the integrating discs and consequently more accurate results than with the fixed point.

When tracing the diagram by No. 2 point, hold the tracing arm close to the fixed point but follow with the eye the movements of No. 2 point.

When using the fixed point, fake off the sliding point No. 2.

### Values

 In the following formulae the travel of roller A is denoted by a " " " M " " " m " " " I " " " i

When the fixed tracing point has been used, then:

### For the English Instrument

 Area A = 0.02 a in.2 Moment M = 0.04 m in.3 Moment of inertia I = 0.32 a − 0.1 i in.4

### For the Metric Instrument

 Area A = 0.12 a cm2 Moment M = 0.6 m cm3 Moment of inertia I = 4 (3 a − i) cm4 Fig. 2

### Example 1 :

Given a circle of respectively 4 in. and 10 cm diameter. To measure: The area of the circle, the moment and the moment of inertia in reference to the tangent x – x.

Place the rail as shown in the adjoining figure, the circle being inside i.e., above the axis of moments. Circumscribe the circle in the way explained, Thus, using Amsler Integrator No. 1, for the fixed tracing point:

### English Instrument

 A disc M disc I disc Initial reading 1832 628 4495 628 3721 1381 Final reading 2460 5123 5102 a = 628 m = 628 i = 1381 Area A = 0.02 × 628 = 12.56 sq. in. Moment M = 0.04 × 628 = 25.12 in. × sq.in. Moment of Inertia I = 0.32 × 628 − 0.1 × 1381 = 62.86 sq. in. × sq. in.

### Metric Instrument

 A disc M disc I disc Initial reading 3271 654 1427 654 8843 1350 Final reading 3925 2081 10193 a = 654 m = 654 i = 1350 Area A = 0.12 × 654 = 78.5 cm2 Moment M = 0.6 × 654 = 392.4 cm × cm2 Moment of Inertia I = 12 × 654 − 4 × 1350 = 2450 cm × cm3

### Example 2:

Take as axis of moments the fop tangent of the same circle and repeat the measurement starting from the same first readings as before. Then:

### Metric Instrument

A discM discI disc A disc M discI disc
1822 6284495 −6283721 13813271 6541427−654 88431350
2460 3867 5102 3925 077310193
a = 628 m = −628 i = 1381 a = 654 m = − 654 i = 1350 Fig. 3

This shows that the A and I discs perform the same travel whether the diagram lies inside or outside the axis, but that the M disc in the latter case moves in the opposite direction. The travel must then be taken into account as a negative quantity.

The travel of the A disc is always a forward one. The travel of the M disc is forward or backward — positive or negative — according as to whether the greater portion of the diagram lies inside or outside the axis of moments.
If the travel of the M disc turns out negative, the centre of gravity of the diagram lies outside, i.e. below the axis of moments.

The travel of the I disc is in most cases a forward one. It is only negative if the whole diagram lies far off the axis of moments. In such cases the I disc may turn backward, then its travel i must be taken as negative and the second member in the formulae for I must accordingly, now be added instead of subtracted.

Before or after an exact measurement roughly circumscribe the diagram while watching the decadic drums of the discs to ascertain the direction and the approximate of the travel of each disc.

Each measurement claiming reliability ought to be gone over at least twice.

### No. 2 Tracing point

For the No. 2 tracing point, the following formulae must be used:

### For the English Instrument

 Area A = 0.01 a in.2 Moment M = 0.01 m in.3 Moment of inertia I = 0.32 a − 0.1 i 8 in.4

### For the Metric Instrument

 Area A = 0.06 a cm2 Moment M = 0.15 m cm3 Moment of inertia I = ½ (3 a − i) cm4

### Example 3:

Determine the resistance of a rail under cross-bending stress. (The diagram fig. 4 is drawn full size, so that the following results can be obtained again almost exactly by measuring directly on the figure.) Fig. 4(In the original the base is 9 cm long)

Draw the line x — x parallel to the foot line of the rail, so that No. 2 point of the integrator takes in the whole profile, the line x — x being chosen as the axis of moments.

Adjust the rail to the line x — x, as the axis of moments, measure the area and the moment of profile by means of No. 2 tracing point (the moment of inertia is not wanted now). Thus:

### English Instrument

A discM disc

(1)
(2)
9729
0164
0600
435
436
1344
1479
1613
135
134
Means:a = 435.5 m = 134.5
A = 0.01 × 435.5 = 4.355 sq. in.
M = 0.01 × 134.5 = 1.345 in. × sq. in.
 h = M A = 1.345 4.355 = 0.309 in.

### Metric Instrument

A discM disc

(1)
(2)
6296
6766
7234
470
468
7108
7255
7403
147
148
Means:a = 469 m = 147,5
A = 0.06 × 469 = 28,15 cm2
M = 0.15 × 147.5 = 22,12 cm3
 h = M A = 22,12 28,15 = 0,786 cm

h being the height of centre of gravity of profile above axis x — x.
(if m were negative, then the M and h would be negative too, and xn — xn would lie below x — x.)

Set off the neutral axis xn — xn, adjust the rail to the line xn — xn as the new axis of moments, and again circumscribe the profile by means of No. 2 tracing point, taking readings on all the discs. Thus:

### English Instrument

 A disc M disc I disc (1)(2) 080012351671 435436 258425842583 0−1 143620882740 652652 Mean values: a = 435.5 m = − 0.05 i = 652

### Metric Instrument

 A disc M disc I disc (1)(2) 795884288896 470468 773077287728 −2  0 676173928022 631630 Mean values: a = 469 m = − 1 i = 630.5
(The measurements of a and m are to check the foregoing operations,)

### English Instrument

 Moment of inertia In = 0.32 × 435.5 − 0.1 × 652   8 = 9.27 sq. in. × sq. in.
Distance of extreme fibre of rail from neutral axis z = 2.06 in.
 Moment of resistance W = In——z = 9.27————2.06 = 4.50 in. × sq. in.

### Metric Instrument

 Moment of inertia In = 3 × 469 − 630.5   2 = 388.25 cm4
Distance of extreme fibre of rail from neutral axis z = 5,22 cm
 Moment of resistance W = In——z = 388.5—————5.22 = 74.4 cm3

### Example 4:

Suppose the diagram is so far off the axis of moments xo — xo (shown in the adjoining figure), that the tracing point cannot reach the whole outline when the integrator is adjusted to xo — xo. Fig. 5

Then draw another parallel axis x — x across the diagram, so that the whole diagram now falls within the reach of the tracing point when the integrator is set to x — x. Adjust the integrator to axis x — x and determine the area and the position of the neutral axis xn — xn (line through centre of gravity of area parallel to xo — xo).

Adjust now the integrator to axis xn — xn and measure the moment of inertia In about axis xn — xn.

If e expresses (in inches or cm) the distance between the lines xo — xo and xn — xn, A the area of the diagram, Mo the moment and Mo the moment of inertia about the axis xo — xo, then:

Mo = e A       Io = In + e2 A

If would be possible, but not advisable, to determine Mo and Io directly from the measurement about axis x — x.

### Large Figures

Large figures exceeding the range of the tracing point must be cut into smaller portions. If it is found impossible to follow the single portions with the tracing point after setting the rail of the Integrator to the axis of moments, an auxiliary axis parallel to the original axis of moments must be drawn for each portion.

Then determine in each portion of the figure the neutral axis parallel to the original axis, the moment of inertia about the neutral axis and the area, and deduce the moment and the moment of inertia of each portion about the original axis by means of the formulae: M = e A       I = In + e2 A

Lastly, sum up the areas A of the single portions, the moments M, and the moments of inertia I.

The above sums will represent the area and the moments of the whole figure.

Very long but narrow figures as, for example, the plan of the water lines of a ship, cause but little trouble; they may simply be intersected by a set of lines at right angles to the axis of moments. The sum of the moments of the portions will then make up the moment of the whole figure.

### Position of centre of gravity of an area

Draw across the figure any two lines x — x and y — y, approximately at right angles to one another. Measure the area A and the Moment Mx about axis x — x, and afterwards the moment My about axis y — y. Mx⁄A will then be the distance of the line xn — xn from the parallel axis x — x and My⁄A the distance of yn — yn from y — y. The point C of intersection of xn — xn and yn — yn will now be the centre of gravity of the area. Fig. 6

### Scales

The foregoing formulae apply to measurements on full-size drawings. If the scale of the drawing be n in. = 1 ft., respectively 1 cm = n cm, the following formulae are to be used:

For the fixed tracing point

### English Instrument

Area A =
 0.02 a × 1 n2
sq. ft.
Moment M =
 0.04 m × 1 n3
cb. ft.
Moment of inertia I =
 (0.32 a − 0.1 i) × 1 n4
sq. × sq. ft.

### Metric Instrument

 Area A = 0.12 a × n2 cm2 Moment M = 0.6 m × n3 cm3 Moment of inertia I = 4 (3 a − i) × n4 cm4

For No. 2 tracing point

### English Instrument

Area A =
 0.01 a × 1 n2
sq. ft.
Moment M =
 0.01 m × 1 n3
cb. ft.
Moment of inertia I =
 (0.32 a − 0.1 i) 8 × 1 n4
sq. × sq. ft.

### Metric Instrument

 Area A = 0.06 a × n2 cm2 Moment M = 0.15 m × n3 cm3 Moment of inertia I = ½ (3 a − i) × n4 cm4

### Example 4:

If the scale be ¼ in. = 1 ft., then for the English Instrument
 n = 1 4 , make 1 n = 4, 1 n2 = 16, 1 n3 = 64, 1 n4 = 256.

For the metric instrument the unit of measurement for A, M, I, again is the centimetre. If, for instance, in the earlier example of the rail the scale had been 1 : 2, the preceeding formulae would have been used, and as now 1 : n = 1 :2, make n = 2. As n2 = 4, n3 = 8, n4 = 16, the formulae, in this particular case are A = 0.24 a, M = 1.2 m, I = 24 a − 8 i, for the movable tracing point No. 2 used for the measurements.

If it is desired to obtain the results in metres instead of centimetres, as is usual in shipbulding, make use of the following formulae

For the fixed tracing point For No. 2 tracing point
A =
 0.12 a ( n  100 ) 2
A =
 0.06 a ( n  100 ) 2
M =
 0.6 m ( n  100 ) 3
M =
 0.15 m ( n  100 ) 3
I =
 4 (3 a - i) ( n  100 ) 4
I =
 ½ (3 a - i) ( n  100 ) 4

### Example 5:

If the scale of the drawing to be evaluated be 1:50, then n = 50, viz.:
 n  100 = 1 2 ; ( n  100 ) 2 = 1 2 ; ( n  100 ) 3 = 1 8 ; ( n  100 ) 4 = 1 16 ;
and the above formulae are for this particular case as follows:

Fixed tracing point

No. 2 tracing point
A =
 0.12 a4 sq. m.
A =
 0.06 a4 sq. m.
M =
 0.6 m8 cb. m.
M =
 0.15 m8 cb. m.
I =
 4 (3 a − i)16 sq. × sq. m.
A =
 1⁄32(3 a − i) sq. × sq. m.

### General remarks on Mechanism

Avoid touching the rims of the integrating discs. They are liable to be spoiled by rust.

Do not try to set the discs to zero; this would involve more time and trouble than taking the readings as they stand.

The pivot centres should occasionally be lubricated with fine oil.

Any alterations of the tracing arm or axial sliding of the wheels of the carriage disturb the adjustment of the integrator and should therefore be avoided.

Do not polish and avoid scratching the integrating discs.

Protect the pivots of the disc shafts from becoming damaged, because the use of the Integrator is very dependent on this point.

Should the adjustment of the Integrator be slightly disturbed, it is nevertheless possible to get good results by going over the measurements twice, first in the usual way and the second time by turning the drawing upside down, so that part of the figure which was first situated between the axis of moments and the steel rail the second time lies below the axis of moments. The average of both measurements should then be taken.

The supplement for the Type 627 integrator: # integrators

## type 627

 With 3 measuring units suitable for evaluating 1. the Area ∫ y dx = A 2. the Static Moment 1⁄2 ∫ y2 dx = M 3. the Moment of Inertia 1⁄3 ∫ y3 dx = I

Supplementary instructions

The Amsler Integrator Type 627 is of the same construction as Type 626, but of larger size, suitable for evaluating larger figures.

The tracing arm, in addition to the fixed tracing point, has two movable tracing points, No. 2 and No. 3.

All the formulae and examples for the fixed tracing point given in the instructions how to use the Amsler Integrator Type 626 apply unchanged to the tracing point No. 2 of the Integrator Type 627, and those for the movable tracing point No. 2 of the Integrator Type 626, apply to the tracing point No. 3 of the Integrator Type 627.

The corresponding values and formulae for the fixed tracing point of the Integrator Type 627 are:

### Values:

 English Instrument Metric Instrument Area A = 0.04 a   in.2 0.24 a  cm2 Moment M = 0.16 m   in.3 2.4 m  cm3 Moment of Inertia I = 2.56 a − 0.8  in.4 32 (3 a − i)   cm4

### Scales:

The foregoing formulae apply to measurements on full-size drawings. If the scale of the drawing be n in. = 1 ft., respectively 1 cm = n cm, the following formulae are to be used:

English Instrument Metric Instrument
Area A =
 0.04 a × 1 n2 sq. ft.
0.24 a × n2 cm2
Moment M =
 0.16 m × 1 n3 cb. ft.
2.4 m × n3 cm3
Moment of Inertia I =
 (2.56 a − 0.8 i) × 1 n4 sq. ft. × sq. ft.
32 (3 a − i) × n4 cm4

If it is desired to obtain the results in metres instead of centimetres, make use of the following formulae

Area A =   0.24 a
 ( n  100 ) 2
Moment M =   2.4 m
 ( n  100 ) 3
Moment of Inertia I =   32 ( 3 a − i )
 ( n  100 ) 4

One of two typewritten notes nailed to the Integrator's box:

## Amsler Integrators 627 and 628

In order to get a sufficient accuracy a substantial part of the measured figure should have a distance from the axis of moments of
 at least 6" when using N°. 1 tracing point at least 3" when using N°. 2 tracing point at least 1½" when using N°. 3 tracing point A.J. Amsler & Co.

The other typewritten note nailed to the Integrator's box:

## Important

Handle the instrument with care. When removing it from the case and placing it on the guide rail, grasp it with the left hand on the base plate and with the right hand on the tracing arm. Introduce the guide rollers into the groove of the guide rail and only afterwards gently lower the front part of the instrument with the integrating rollers on the tracing board.

Under no circumstances displace the guide rollers in their axial direction or the tracing points from the positions fixed by the makers. Do not disturb the adjustment of the integrating rollers and counting mechanisms. Take care that the gauges are not bent.