Hello everyone,
I've bought a "cheap" but heavily blue-white-green corroded HP-25c and washed and cleaned it. It was so damaged that in the wash process I almost lost a component; this the "big" black transistor on the left of the first (mid) RAM chip (thank you, Harald!):
Now, I have to test it (and every component I guess) and to solder it (or a
new one) on the board again. But where does it connect onto the board? I didn't find any picture or schematic neither in this site, nor on the PPC DVD of Jack Schwartz.
Any help will be appreciated.
[
attachment=6268]I've attached an ancient PDF that I had lurking around - hope that might help?
I've also got a NASA HP-25 program listing for calculating Loran time differences if you want to run and see if it works ok after your major trauma surgery on the calculator.
Dennis
If you want to see the last photo in whole go to the website that is shown in the pdf. So you can see the whole hardware displayed.
Andi
(08-31-2018 07:53 PM)SlideRule Wrote: [ -> ]Can be downloaded from archive.org as NTRS-1978002312 LORAN Time Difference.
Thanks for the link to the program.
These formulas are used:
\(
\begin{align*}
\beta_T&=\tan^{-1}[C\tan(\Phi_T) ] \\
\beta_R&=\tan^{-1}[C\tan(\Phi_R) ] \\
X &= \cos^{-1}[\sin(\beta_R)\sin(\beta_T)+\cos(\beta_R)\cos(\beta_T)\cos(\lambda_R-\lambda_T)]
\end{align*}
\)
But we can factor out \(\cos(\beta_R)\cos(\beta_T)\) and get:
\(X = \cos^{-1}[\cos(\beta_R)\cos(\beta_T)(\tan(\beta_R)\tan(\beta_T)+\cos(\lambda_R-\lambda_T))]\)
Hence we can avoid calculating \(\sin(\beta_R)\) and \(\sin(\beta_T)\) since \(\tan(\beta_R)\) and \(\tan(\beta_T)\) have already been computed.
This makes the program a bit shorter:
Code:
01: 24 06 RCL 6 C
02: 24 01 RCL 1 Φ_T C
03: 14 06 f tan tan(Φ_T) C
04: 61 × C tan(Φ_T) = tan(β_T)
05: 31 ENTER tan(β_T) tan(β_T)
06: 15 06 g tan⁻¹ β_T tan(β_T)
07: 14 05 f cos cos(β_T) tan(β_T)
08: 24 06 RCL 6 C cos(β_T) tan(β_T)
09: 24 03 RCL 3 Φ_R C cos(β_T) tan(β_T)
10: 14 06 f tan tan(Φ_R) C cos(β_T) tan(β_T)
11: 61 × C tan(Φ_R) cos(β_T) tan(β_T) tan(β_T)
12: 31 ENTER tan(β_R) tan(β_R) cos(β_T) tan(β_T)
13: 22 R↓ tan(β_R) cos(β_T) tan(β_T) tan(β_R)
14: 15 06 g tan⁻¹ β_R cos(β_T) tan(β_T) tan(β_R)
15: 14 05 f cos cos(β_R) cos(β_T) tan(β_T) tan(β_R)
16: 61 × cos(β_R)cos(β_T) tan(β_T) tan(β_R) tan(β_R)
17: 22 R↓ tan(β_T) tan(β_R) tan(β_R) cos(β_R)cos(β_T)
18: 61 × tan(β_R)tan(β_T) tan(β_R) cos(β_R)cos(β_T) cos(β_R)cos(β_T)
19: 21 x<>y tan(β_R) tan(β_R)tan(β_T) cos(β_R)cos(β_T) cos(β_R)cos(β_T)
20: 22 R↓ tan(β_R)tan(β_T) cos(β_R)cos(β_T) cos(β_R)cos(β_T) tan(β_R)
21: 24 04 RCL 4 λ_R tan(β_R)tan(β_T) cos(β_R)cos(β_T) cos(β_R)cos(β_T)
22: 24 02 RCL 2 λ_T λ_R tan(β_R)tan(β_T) cos(β_R)cos(β_T)
23: 41 - λ_R-λ_T tan(β_R)tan(β_T) cos(β_R)cos(β_T)
24: 14 05 f cos cos(λ_R-λ_T) tan(β_R)tan(β_T) cos(β_R)cos(β_T)
25: 51 + tan(β_R)tan(β_T)+cos(λ_R-λ_T) cos(β_R)cos(β_T)
26: 61 x cos(β_R)cos(β_T)(tan(β_R)tan(β_T)+cos(λ_R-λ_T)) = cos(X)
27: 15 05 g cos⁻¹ X
28: 24 05 RCL 5 A(rad) X
29: 61 × d = AX
30: 74 R/S d
31: 24 00 RCL 0 T_m T_s
32: 41 - T_s-T_m
Example
21282.339
π
×
180
÷
STO 5
0.99664767
STO 6
39.1930
→H
STO 3
82.0615
→H
STO 4
34.034604
→H
STO 1
77.544676
→H
STO 2
f CLEAR PRGM
R/S
2317.7679
Cheers
Thomas
(08-31-2018 01:53 PM)Leviset Wrote: [ -> ]I've attached an ancient PDF that I had lurking around - hope that might help?
I've also got a NASA HP-25 program listing for calculating Loran time differences if you want to run and see if it works ok after your major trauma surgery on the calculator.
Dennis
Sadly, Jaques’ site, while excellent, is on the 21 and 25, but not the 25c. While there is much relevant information, the original question seems to still be unanswered: Is there a schematic for the 25c available? Thanks.-kby
(06-17-2019 06:40 PM)AndiGer Wrote: [ -> ]Did you look at Tony's schematics? HP25C included ...
https://www.hpmuseum.org/forum/thread-13065.html
Mea culpa. I had, but apparently not carefully enough. I think maybe I got an earlier version, or the name "Classic Schematics" stuck in my mind, although I did know he subsequently added the 19c and 29c.-kby