This manual came with a Amsler Integrator 627/499, sold by Keuffel & Esser under number 3078S.
Apart from the manual, there were a supplement and two typewritten notes.
With 3 measuring units suitable for evaluating  
1.  the Area  ∫  y  dx = A 
2.  the Static Moment  ^{1}⁄_{2} ∫  y^{2}  dx = M 
3.  the Moment of Inertia  ^{1}⁄_{3} ∫  y^{3}  dx = I 
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.
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.
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.
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:
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} 
Area  A =  0.12 a  cm^{2}  
Moment  M =  0.6 m  cm^{3}  
Moment of inertia  I =  4 (3 a − i)  cm^{4} 
Fig. 2 
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:
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. 
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 cm^{2}  
Moment  M  = 0.6 × 654 = 392.4 cm × cm^{2}  
Moment of Inertia  I  = 12 × 654 − 4 × 1350 = 2450 cm × cm^{3} 
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:
English Instrument  Metric Instrument  
A disc  M disc  I disc  A disc  M disc  I disc  
1822  628  4495  −628  3721  1381  3271  654  1427  −654  8843  1350  
2460  3867  5102  3925  0773  10193  
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.
For the No. 2 tracing point, the following formulae must be used:
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} 
Area  A =  0.06 a  cm^{2}  
Moment  M =  0.15 m  cm^{3}  
Moment of inertia  I =  ½ (3 a − i)  cm^{4} 
Determine the resistance of a rail under crossbending 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:
A disc  M 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.  

A disc  M 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 cm^{2}  
M = 0.15 × 147.5 = 22,12 cm^{3}  

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
x_{n} — x_{n}
would lie below x — x.)
Set off the neutral axis x_{n} — x_{n}, adjust the rail to the line x_{n} — x_{n} 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:
A disc  M disc  I disc  
(1) (2)  0800 1235 1671  435 436  2584 2584 2583  0 −1  1436 2088 2740  652 652 
Mean values:  a = 435.5  m = − 0.05  i = 652 
A disc  M disc  I disc  
(1) (2)  7958 8428 8896  470 468  7730 7728 7728  −2 0  6761 7392 8022  631 630 
Mean values:  a = 469  m = − 1  i = 630.5 
Moment of inertia  I_{n} =  0.32 × 435.5 − 0.1 × 652 8  = 9.27 sq. in. × sq. in. 
Moment of resistance  W =  I_{n} —— z  =  9.27 ———— 2.06  = 4.50 in. × sq. in. 
Moment of inertia  I_{n} =  3 × 469 − 630.5 2  = 388.25 cm^{4} 
Moment of resistance  W =  I_{n} —— z  =  388.5 ————— 5.22  = 74.4 cm^{3} 
Suppose the diagram is so far off the axis of moments x_{o} — x_{o} (shown in the adjoining figure), that the tracing point cannot reach the whole outline when the integrator is adjusted to x_{o} — x_{o}.
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 x_{n} — x_{n} (line through centre of gravity of area parallel to x_{o} — x_{o}).
Adjust now the integrator to axis x_{n} — x_{n} and measure the moment of inertia I_{n} about axis x_{n} — x_{n}.
If e expresses (in inches or cm) the distance between the lines x_{o} — x_{o} and x_{n} — x_{n}, A the area of the diagram, M_{o} the moment and M_{o} the moment of inertia about the axis x_{o} — x_{o}, then:
M_{o} = e A I_{o} = I_{n} + e^{2} A
If would be possible, but not advisable, to determine M_{o} and I_{o} directly from the measurement about axis x — x.
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 = I_{n} + e^{2} 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.
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 M_{x} about axis x — x, and afterwards the moment M_{y} about axis y — y. M_{x}⁄A will then be the distance of the line x_{n} — x_{n} from the parallel axis x — x and M_{y}⁄A the distance of y_{n} — y_{n} from y — y. The point C of intersection of x_{n} — x_{n} and y_{n} — y_{n} will now be the centre of gravity of the area.
Fig. 6 
The foregoing formulae apply to measurements on fullsize 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
Area  A = 
 sq. ft.  
Moment  M = 
 cb. ft.  
Moment of inertia  I = 
 sq. × sq. ft. 
Area  A =  0.12 a × n^{2}  cm^{2}  
Moment  M =  0.6 m × n^{3}  cm^{3}  
Moment of inertia  I =  4 (3 a − i) × n^{4}  cm^{4} 
For No. 2 tracing point
Area  A = 
 sq. ft.  
Moment  M = 
 cb. ft.  
Moment of inertia  I = 
 sq. × sq. ft. 
Area  A =  0.06 a × n^{2}  cm^{2}  
Moment  M =  0.15 m × n^{3}  cm^{3}  
Moment of inertia  I =  ½ (3 a − i) × n^{4}  cm^{4} 
n =  1 4  , make  1 n  = 4,  1 n^{2}  = 16,  1 n^{3}  = 64,  1 n^{4}  = 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 n^{2} = 4, n^{3} = 8, n^{4} = 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 = 

A = 
 
M = 

M = 
 
I = 

I = 

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  ; 
Fixed tracing point  No. 2 tracing point  
A = 

A = 
 
M = 

M = 
 
I = 

A = 

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:
With 3 measuring units suitable for evaluating  
1.  the Area  ∫  y  dx = A 
2.  the Static Moment  ^{1}⁄_{2} ∫  y^{2}  dx = M 
3.  the Moment of Inertia  ^{1}⁄_{3} ∫  y^{3}  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:
English Instrument  Metric Instrument  
Area  A =  0.04 a in.^{2}  0.24 a cm^{2}  
Moment  M =  0.16 m in.^{3}  2.4 m cm^{3}  
Moment of Inertia  I =  2.56 a − 0.8 in.^{4}  32 (3 a − i) cm^{4} 
The foregoing formulae apply to measurements on fullsize 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 = 
 0.24 a × n^{2} cm^{2}  
Moment  M = 
 2.4 m × n^{3} cm^{3}  
Moment of Inertia  I = 
 32 (3 a − i) × n^{4} cm^{4} 
If it is desired to obtain the results in metres instead of centimetres, make use of the following formulae
Area  A =  0.24 a 
 
Moment  M =  2.4 m 
 
Moment of Inertia  I =  32 ( 3 a − i ) 

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