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Old 12-23-2015, 08:48 AM   #1
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MGMike .38
Specifications, hammer block spring

I found a Walther factory blueprint for the hammer block spring of a PP-series pistol. This little spring does double duty: it holds the hammer block in its "blocking" position except when the trigger is pulled AND it pushes the sear (the stirrup-shaped "cocking piece" into engagement with the hammer to hold the hammer at full cock for each shot. The latter must be accomplished during the brief interval when the hammer is rotated rearward by the slide and the hammer's toe swings into alignment with a horizontal groove on the sear's rear face. Thus it is a critical spring. It assures engagement of the hammer block in case the pistol is dropped on its hammer; it keeps the hammer from following the slide back to battery, and it precludes the pistol firing uncontrollably in full automatic.

Unfortunately the spring is very small and difficult to assess. Also, once removed it easily confused with other springs taken out of the pistol. On the other hand it has the virtue of being readily accessible simply by depressing and tilting out the spring plug in the top of the frame, visible when the slide is dismounted.

For those with micrometers, here are the specifications. It is Walther part No. 221 77 67, current as of 1986.

OAL: 0.393" (10mm)
Coil OD: .098" (2.5mm)
Active coils: 9
Total coils: 11
Wire diameter: .015" (0.4mm)

The ends are closed and ground flat. One end is visibly reduced in diameter, which helps to identify it. This end fits snugly over the post on the hammer block to keep the spring from being lost in disassembly. The wire material is C DIN 2076-II, normalized after forming.

According to the drawing, the spring rate is c= 319 p/mm. I'm not smart enough to figure out what that means, but there are bound to be some engineers here who can interpret it.

M
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Old 12-23-2015, 01:18 PM   #2
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Thank you so much Mike. Especially on how important this tiny spring relates to the cocking piece & hammer.
One thing I'm a bit confused on is: On the bottom base of the spring plug post, there is a groove cut that locks on the smaller dia of the spring, so during disassembly, the plug and the spring come out as a unit. The spring end is tight going on the post, but snaps in the groove when it reaches it. All the Walthers I've taken apart the plug and spring always came out as a unit.
I'm just wondering. Thanks!
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Old 12-23-2015, 04:35 PM   #3
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colt .22
MGMike

Thank you for the information. I am having trouble finding a gunsmith in OKC that knows anything about Walthers.I am still asking around to see if anyone I know have taken a PPK down for repair.Thank you again for your help.
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Old 12-23-2015, 05:17 PM   #4
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Originally Posted by ViperR View Post
...
One thing I'm a bit confused on is: On the bottom base of the spring plug post, there is a groove cut that locks on the smaller dia of the spring, so during disassembly, the plug and the spring come out as a unit. The spring end is tight going on the post, but snaps in the groove when it reaches it. All the Walthers I've taken apart the plug and spring always came out as a unit.
...
Viper, you'd make a good diplomat. I think you are gently trying to correct me.

You are not confused at all; in fact you are correct. Writing from memory alone (a poor practice) I had it exactly backward. The small end of the spring snaps onto the plug, not onto the hammer block.

Thank you.

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Old 12-23-2015, 09:10 PM   #5
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And you are a gentleman and a valued member of the Walther community in our joint venture to educate all who are in need of guidance.
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Old 02-02-2018, 09:09 AM   #6
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Originally Posted by MGMike View Post
For those with micrometers, here are the specifications. It is Walther part No. 221 77 67, current as of 1986.

OAL: 0.393" (10mm)
Coil OD: .098" (2.5mm)
Active coils: 9
Total coils: 11
Wire diameter: .015" (0.4mm)

The ends are closed and ground flat. One end is visibly reduced in diameter, which helps to identify it. This end fits snugly over the post on the hammer block to keep the spring from being lost in disassembly. The wire material is C DIN 2076-II, normalized after forming.

According to the drawing, the spring rate is c= 319 p/mm. I'm not smart enough to figure out what that means, but there are bound to be some engineers here who can interpret it.

M
Bit of an old thread but if anyone is still interested I did some some digging and worked out the spring rate translates to 3.13N/mm in modern units.

Practically speaking that means 4lbs weight is required to compress it fully.
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Old 10-30-2018, 05:01 PM   #7
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macpenna .22
hammer block spring

Earl Sheehen , the Walther distributor, has hammer block springs. He also repairs Walther pistols

Last edited by macpenna; 10-30-2018 at 05:05 PM.
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Old 10-30-2018, 05:42 PM   #8
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...former Walther distributor. The distributor for Carl Walther GmbH Deutschland is Walther Arms LLC, Fort Smith, Arkansas.
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Old 10-31-2018, 11:12 AM   #9
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Early ppk

Mr. Sheehen says he is a distributor. Just had my ppk repaired with new hammer block spring, and my ppk is circa 1931. Those are hard to find anywhere else. Local gunsmith had my ppk for a year because he could not find the part for it.
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Old 03-18-2019, 05:18 PM   #10
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Originally Posted by JamesWilkinson View Post
Bit of an old thread but if anyone is still interested I did some some digging and worked out the spring rate translates to 3.13N/mm in modern units.

Practically speaking that means 4lbs weight is required to compress it fully.
OK...I'l bite. I'm a Mechanical Engineer and I never heard of a spring rate given in the original terms stated. If you indeed found out what it means, can you identify the units used? Always learning.
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