Counting of Reed Count and Heald Count in Weaving
Counting of reed count:
Reed determines the spacing of warp threads. Ordinary reed used for weaving cotton fabrics is made of rustless steel wires. They are bound at each end on baulks by pith bands, whose thickness determines the spacing of dents. Sometimes brass wires are used instead of rustless steel.
Required Count,
Dents Per inch in known system
= …………………………………………….. × count in known system
Dents Per inch in required system
The following two tables will be used during counting Reed & Heald count in weaving.
System based on the number of dents in a given space.
Dents Per inch in known system
= …………………………………………….. × count in known system
Dents Per inch in required system
The following two tables will be used during counting Reed & Heald count in weaving.
System based on the number of dents in a given space.
|
Name
of System
|
Basis
Of numbering
|
|
Stockport
|
Number
Of dents per 2 inches
|
|
Radcliff
|
Number
Of dents per 1 inch.
|
|
Huddersfield
|
Number
Of dents per 1 inch.
|
|
Metric
|
Number
Of dents per 1 decm.
|
System based on the number of groups or beers in a given space.
|
Name
of System
|
Basis
Of numbering
|
|
Bolton
|
20
dents per 24.5 inches.
|
|
Bradford
|
20
dents per 36 inches.
|
|
Blackburn
|
20
dents per 45 inches.
|
|
Irish
|
100
dents per 40 inches.
|
|
Leeds
|
19
dents per 9 inches.
|
|
Macclesfield
|
100
dents per 36 inches.
|
All the above systems are convertible into dents per inch basis.
Example No. 01:
Find the count of 10s Irish reed into the Huddersfield System .
Solution:
10s Irish,
10 × 100
= ……………
40
= 25 dents per inch.
1 Huddersfield = 1 dent per inch.
Therefore,
10s Irish,
25
= ………
1
= 25s Huddersfield.
So, the count in Huddersfield System is 25s
Example No. 02:
Find the count of a 40s Radcliff in the Stockport System.
Solution:
Count,
40 × 2
= …………..
1
= 80s Stockport.
So, the count in Stockport System is 80s
Example No. 03:
Find the number of ends per inch in a reed of 3/72s Bradford.
Solution:
72s Bradford ,
72 × 20
= ………………….. ends per inch
36
= 40 ends per inch.
Therefore, the number of ends per inch = 3 × 4
= 120 ends per inch.
Example No. 01:
Find the count of 10s Irish reed into the Huddersfield System .
Solution:
10s Irish,
10 × 100
= ……………
40
= 25 dents per inch.
1 Huddersfield = 1 dent per inch.
Therefore,
10s Irish,
25
= ………
1
= 25s Huddersfield.
So, the count in Huddersfield System is 25s
Example No. 02:
Find the count of a 40s Radcliff in the Stockport System.
Solution:
Count,
40 × 2
= …………..
1
= 80s Stockport.
So, the count in Stockport System is 80s
Example No. 03:
Find the number of ends per inch in a reed of 3/72s Bradford.
Solution:
72s Bradford ,
72 × 20
= ………………….. ends per inch
36
= 40 ends per inch.
Therefore, the number of ends per inch = 3 × 4
= 120 ends per inch.
Counting of heald count:
The number of heald eyes per inch across
the healds in a set expresses the count of the healds. When a set
contains 4 shafts, it is called a plain set.
Example No. 04:
Find the count of the healds that will be required for weaving a 6 shaft satin fabric using 72s Stockport reed, drawn 3 ends per dent.
Solution:
Number of ends per inch in the reed,
3 × 72
= ………………..
2
= 108 ends.
Therefore, rate of knitting,
108
= …………..
6
= 18 healds per inch on the rib band.
Thus we require 6 healds of 72s plain set.
Example No. 04:
Find the count of the healds that will be required for weaving a 6 shaft satin fabric using 72s Stockport reed, drawn 3 ends per dent.
Solution:
Number of ends per inch in the reed,
3 × 72
= ………………..
2
= 108 ends.
Therefore, rate of knitting,
108
= …………..
6
= 18 healds per inch on the rib band.
Thus we require 6 healds of 72s plain set.
