(m)
Install the spinner front and rear cones (AMM 72-31-01/401).
S 974-005-C00
(4) Do the imbalance correction.
(a)
Calculate the delta between the moment-weight differences of the removed blade pair and the spare blade pair. 1) If the delta between the moment-weight differences of the
matched pairs is less than 50.0 gram-inches (127.0 gm-cm), no correction is necessary.
(b)
If more than one pair of replacement blades are installed, do a vector diagram and calculate the resultant, and refer to Figure 401. 1) Figure 401 uses the resultant as the criteria, which could
be an important factor if several pairs of blades are replaced.
(c)
Obey these steps when you do the imbalance correction: 1) It is recommended that you do not put more than one set of balance screws on the spinner.
2) More than one set of balance screw can make future imbalance corrections harder.
(d)
To do the imbalance correction you must make a vector diagram to find the total sum of all the corrections.
(e)
An example of this procedure is as follows (Fig. 403):
1) For this example, you removed these blades with these moment-weights: No. 1: 31849 gram-inches (80896.46 gm-cm) No. 38: 32505 gram-inches (82562.7 gm-cm) No. 3: 30910 gram-inches (78511.4 gm-cm) No. 35: 31340 gram-inches (79603.6 gm-cm)
2) The moment-weights of the opposite blades are as follows:
No. 20: 31804 gram-inches (80782.16 gm-cm) No. 19: 32545 gram-inches (82664.3 gm-cm) No. 22: 30940 gram-inches (78587.6 gm-cm) No. 16: 31335 gram-inches (79590.9 gm-cm)
3) Calculate the differences between the moment-weights: D1: (No. 1/20) = 31849 - 31804 = 45 gram-inches
(114.3 gm-cm) D2: (No. 19/38) = 32545 - 32505 = 40 gram-inches
(101.6 gm-cm) D3: (No. 22/3) = 30940 - 30910 = 30 gram-inches
(76.2 gm-cm) D4: (No. 35/16) = 31340 - 31335 = 5 gram-inches
(12.7 gm-cm)
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ALL ú ú C03 Page 417 ú Mar 12/01
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Calculate the moment-weights differences for the spare fan blades: D'1: (No. 1/20) = 31445 - 31405 = 40 gram-inches
(101.6 gm-cm) D'2: (No. 19/38) = 31628 - 31623 = 5 gram-inches
(12.7
gm-cm) D'3: (No. 22/3) = 32735 - 32730 = 5 gram-inches
(12.7
gm-cm) D'4: (No. 35/16) = 30348 - 30303 = 45 gram-inches
(114.3 gm-cm)
Calculate the differences between the blades that you removed and the spare blades: D1 - D'1 = 45 - 40 = 5 gram-inches (12.7 gm-cm) = F1 D2 - D'2 = 40 -5 = 35 gram-inches (88.9 gm-cm) = F2 D3 - D'3 = 30 -5 = 25 gram-inches (63.5 gm-cm) = F3 D'4 - D4 = 45 -5 = 40 gram-inches (101.6 gm-cm) = F4
NOTE: The balance weight is always positive.
____
Find the direction of the corrections as follows:
NOTE: When D is greater than D', add the balance weight
____ near the heavier blade. When D is lighter than D', add the balance weight near the lighter blade.
a) For the blades No. 1 and 20, D1 is more than D'1. Add the balance weight near the heavier blade: No. 1. b) For the blades No. 38 and 19, D2 is more than D'2. Add the balance weight near the heavier blade: No. 19. c) For the blades No. 3 and 22, D3 is more than D'3. add the balance weight near the heavier blade: No. 22. d) For the blades No. 35 and 16, D4 is less than D'4. Add
the balance weight near the lighter blade: No. 16.Make the vectorial diagram as follows (Fig. 403): a) Use a polar chart and draw the vectors of the
moment-weight differences between the removed blades
and the spare blades.
1.
Use the direction of the correction vector as given
_
above.
2.
The resultant vector FR will show the necessary
_ correction from F1, F2, F3 and F4. b) From the end of the vector F1, make an arc equal to F2 and parallel to F2. c) From the end of vector F2, make an arc equal to F1 and parallel to F1 until the arcs intersect.
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