Part 6 (1/2)
+---------------------+---------------------+--------------------------+ | Given | To Find | Rule | +---------------------+---------------------+--------------------------+ |The taper per foot. |The taper per inch. |Divide the taper per foot | | | | by 12. | | | | | |The taper per inch. |The taper per foot. |Multiply the taper per | | | | inch by 12. | | | | | |End diameters and |The taper per foot. |Subtract small diameter | | length of taper in | | from large; divide by | | inches. | | length of taper, and | | | | multiply quotient by 12.| | | | | |Large diameter and |Diameter at small |Divide taper per foot by | | length of taper in | end in inches. | 12; multiply by length | | inches and taper | | of length of taper, and | | per foot. | | subtract result from | | | | large diameter. | | | | | |Small diameter and |Diameter at large |Divide taper per foot by | | length of taper in | end in inches. | 12; multiply by length | | inches, and taper | | of taper, and add result| | per foot. | | to small diameter. | | | | | |The taper per foot |Distance between | Subtract small diameter | | and two diameters | two given diameters| from large; divide re- | | in inches. | in inches. | mainder by taper per | | | | foot, and multiply | | | | quotient by 12. | | | | | |The taper per foot. |Amount of taper in | Divide taper per foot by | | | a certain length | 12; multiply by given | | | given in inches. | length of tapered part.| +---------------------+---------------------+--------------------------+
=Accurate Measurement of Angles and Tapers.=--When great accuracy is required in the measurement of angles, or when originating tapers, disks are commonly used. The principle of the disk method of taper measurement is that if two disks of unequal diameters are placed either in contact or a certain distance apart, lines tangent to their peripheries will represent an angle or taper, the degree of which depends upon the diameters of the two disks and the distance between them. The gage shown in Fig. 16, which is a form commonly used for originating tapers or measuring angles accurately, is set by means of disks. This gage consists of two adjustable straight-edges _A_ and _A_{1}_, which are in contact with disks _B_ and _B_{1}_. The angle [alpha] or the taper between the straight-edges depends, of course, upon the diameters of the disks and the center distance _C_, and as these three dimensions can be measured accurately, it is possible to set the gage to a given angle within very close limits. Moreover, if a record of the three dimensions is kept, the exact setting of the gage can be reproduced quickly at any time. The following rules may be used for adjusting a gage of this type.
[Ill.u.s.tration: Fig. 16. Disk Gage for Accurate Measurement of Angles and Tapers]
=To Find Center Distance for a Given Taper.=--When the taper, in inches per foot, is given, to determine center distance _C_. _Rule:_ Divide the taper by 24 and find the angle corresponding to the quotient in a table of tangents; then find the sine corresponding to this angle and divide the difference between the disk diameters by twice the sine.
_Example:_ Gage is to be set to 3/4 inch per foot, and disk diameters are 1.25 and 1.5 inch, respectively. Find the required center distance for the disks.
0.75 ---- = 0.03125.
24
The angle whose tangent is 0.03125 equals 1 degree 47.4 minutes; sin 1 47.4' = 0.03123; 1.50 - 1.25 = 0.25 inch;
0.25 ----------- = 4.002 inches = center distance C.
2 0.03123
=To Find Center Distance for a Given Angle.=--When straight-edges must be set to a given angle [alpha], to determine center distance _C_ between disks of known diameter. _Rule:_ Find the sine of half the angle [alpha] in a table of sines; divide the difference between the disk diameters by double this sine.
_Example:_ If an angle [alpha] of 20 degrees is required, and the disks are 1 and 3 inches in diameter, respectively, find the required center distance _C_.
20 ---- = 10 degrees; sin 10 = 0.17365; 2
3 - 1 ----------- = 5.759 inches = center distance _C_.
2 0.17365
=To Find Angle for Given Taper per Foot.=--When the taper in inches per foot is known, and the corresponding angle [alpha] is required. _Rule:_ Divide the taper in inches per foot by 24; find the angle corresponding to the quotient, in a table of tangents, and double this angle.
_Example:_ What angle [alpha] is equivalent to a taper of 1-1/2 inch per foot?
1.5 --- = 0.0625.
24
The angle whose tangent is 0.0625 equals 3 degrees 35 minutes, nearly; then, 3 deg. 35 min. 2 = 7 deg. 10 min.
=To Find Angle for Given Disk Dimensions.=--When the diameters of the large and small disks and the center distance are given, to determine the angle [alpha]. _Rule:_ Divide the difference between the disk diameters by twice the center distance; find the angle corresponding to the quotient, in a table of sines, and double the angle.
_Example:_ If the disk diameters are 1 and 1.5 inch, respectively, and the center distance is 5 inches, find the included angle [alpha].
1.5 - 1 ------- = 0.05.
2 5
The angle whose sine is 0.05 equals 2 degrees 52 minutes; then, 2 deg.
52 min. 2 = 5 deg. 44 min. = angle [alpha].
[Ill.u.s.tration: Fig. 17. Setting Center Mark in Line with Axis of Lathe Spindle by use of Test Indicator]
[Ill.u.s.tration: Fig. 18. Jig-plate with b.u.t.tons attached, ready for Boring]
=Use of the Center Indicator.--=The center test indicator is used for setting a center-punch mark, the position of which corresponds with the center or axis of the hole to be bored, in alignment with the axis of the lathe spindle. To ill.u.s.trate, if two holes are to be bored, say 5 inches apart, small punch marks having that center-to-center distance would be laid out as accurately as possible. One of these marks would then be set central with the lathe spindle by using a center test indicator as shown in Fig. 17. This indicator has a pointer _A_ the end of which is conical and enters the punch mark. The pointer is held by shank _B_ which is fastened in the toolpost. The joint _C_ by means of which the pointer is held to the shank is universal; that is, it allows the pointer to move in any direction. Now when the part being tested is rotated by running the lathe, if the center-punch mark is not in line with the axes of the lathe spindle, obviously the outer end of pointer _A_ will vibrate, and as joint _C_ is quite close to the inner end, a very slight error in the location of the center-punch mark will cause a perceptible movement of the outer end, as indicated by the dotted lines.