Just out.



#7

The latest version of
HP Solve
has Gene's 10bii+ review and more good stuff.


Edited: 29 Apr 2011, 12:11 p.m.


#8

finally! :-)

#9

I can't wait to see that HP-41 replacement CPU board!


#10

indeed! where can we get that 41 replacement board??


#11

41CL

The first set of the boards are spoken for.


Edited: 29 Apr 2011, 4:27 p.m.


#12

Gene,

I have visited the site but I don't see how to purchase one or get on a waiting list. Any pricing information, etc. available?

Thanks.


#13

Sure, right here:

10bII+


The link through the business section is broken. Might be a while before the HP website is fixed.


#14

Gene, how does a reference to HP's order page for a business calc answer a question about how to buy a third-party replacement processor for an HP-41? Am I missing something?


#15

doh. I mixed up the 10bII+ with the 41CL.

No order page for the 41CL next set yet. Let Monte get the beta units out and in the hands of a group before he worries about taking orders for more.


#16

Thanks for the info Gene.

#17

I also found the "Tweaking the HP42S" article interesting. Specially in light of the discussion of the WP34S keyboard layout. Although I have to say that I have a BIG problem with putting the X<>Y key on a shifted keyplane (as seems to be suggested by the image). I use X<>Y a lot more frequently than I use RDN and personally would switch those two if I could not have both unshifted.


#18

It is interesting...quickly checking through the models, the only ones I could find which had x<>y shifted were the 17BII, 17BII+, 19BII, 28C/S (swap), 65 and 67. (For the 65/67, x<>Y and Roll down would be primary only if programs using the appropriate LBL had not been defined in program memory. Also, it is interesting that the 34C, 37E and 38E/C had x<>y primary, but roll down shifted.

...but of course, the purpose of articles like these are specifically for creating a dialog and I'm grateful that any sort of discussion might take place.

Jake Schwartz


#19

Hi Jake,

Thanks for the article! Actually of the three business calculators you mention (17BII, 17BII+ and 19BII) the X<>Y and RDN keys are printed as if they are shifted but they act as unshifted keys when in RPN mode.

Marwan

#20

Jake's new article has ancestors in 2008 as he mentioned in the last paragraph in HP Solve. Looking at them, you'll find tweaking is an hobby with a long tradition in this community, and also the wp34s has its parents and grandparents :-)

Quote:
I use X<>Y a lot more frequently than I use RDN and personally would switch those two if I could not have both unshifted.

I concur. OTOH only three scientific HPs needed one or both of these functions shifted. Jake listed them.

Walter

#21

Concur, I prefer the swap and the roll down functions (and the STO function) all to be unshifted as well.

The keyboard with three shifted functions (not counting the alpha characters) is a bit crazy.

The key layout with the gold shift on top and blue shift on the bottom look the best IMO.


#22

I'll second that on STO! I guess the logic is that you use RCL more than STO so shift STO. However, I like the old Pioneer layout (and the 41 for that matter) with STO, RCL, X<>Y and RDN all unshifted.

#23

I have to say that Gene's review can be used as a "getting started" guide - better than the one provided with the actual calculator. It's that good!

I say this because it helped me to understand some aspects that I did not "get" from the provided guide at a first reading.


#24

Heh. From your lips to HP's ears.

Seriously, all of that material is in there. I have no idea how the quick start guide was written.

The HP 10bII+ is a great little machine.


#25

Well, the quickstargt guide for the 10bII+ is essentially the "getting started" chapter from the original manual. A lot of the new things were put in there, but there wasn't enough space to cover it all.

The QSG *is* enough for you to effectively use the machine however. Unlike the 30b QSG. :-)

TW

#26

Very good stuff.

#27

Very good stuff indeed, thanks!

I took the time to finally solve the cube resistor problem. I never did all the exercises in my textbooks (quite the contrary!). That one was in Resnick & Halliday's book, but I skipped in '82 :-)


7 o---------R---------o 6
|\ /|
| R R |
| \ 2 1 / |
| o-----R-----o-----------+
| | | | |
R R R R |
| | | | +--+--+
| o-----R-----o | | /|\ | 1A
| / 3 4 \ | | | |
| R R | +--+--+
|/ \| |
0 o---------R---------o 5 |
| |
| |
= =

For easy visualization the cube has been redrawn to 2-D. For convenience, let's use 1 ohm resistors. The circuit will be solved through nodal analysis.
Let's take node 0 (ground) as a reference and inject 1 amp into node 1. For convenience, let R = 1 ohm. The seven nodal equations can be readily obtained:

1) (V1 - V6) + (V1 - V2) + (V1 - V4) = 1
2) (V2 - V7) + (V2 - V1) + (V2 - V3) = 0
3) (V3 - V2) + (V3 - V4) + V3 = 0
4) (V4 - V5) + (V4 - V1) + (V4 - V3) = 0
5) V5 + (V5 - V6) + (V5 - V4) = 0
6) (V6 - V5) + (V6 - V7) + (V6 - V1) = 0
7) (V7 - V6) + (V7 - V2) + V7 = 0

Instead of putting the system of equations into matrix form before solving it, the equations may be used exactly as they are written and solved with help of
the hp-50g program described here. The program finds V1 = .833333333333, which implies the equivalent resistance across the nodes 0 and 1 is 5/6*R. Likewise,
when the 1 Amp current source is connected to the nodes 5 or 6, for instance, other equivalent resistances can be determined. They are 7/12*R and 3/4*R,
respectively.

Sets of equations to be used in the program:

%%HP: T(3)A(D)F(,);
{ 'V1-V6+V1-V2+V1-V4=1,' 'V2-V7+V2-V1+V2-V3=0,' 'V3-V2+V3-V4+V3=0,' 'V4-V5+V4-V1+V4-V3=0,' 'V5+V5-V6+V5-V4=0,' 'V6-V5+V6-V7+V6-V1=0,' 'V7-V6+V7-V2+V7=0,' }

%%HP: T(3)A(D)F(,);
{ 'V1-V6+V1-V2+V1-V4=0,' 'V2-V7+V2-V1+V2-V3=0,' 'V3-V2+V3-V4+V3=0,' 'V4-V5+V4-V1+V4-V3=0,' 'V5+V5-V6+V5-V4=1,' 'V6-V5+V6-V7+V6-V1=0,' 'V7-V6+V7-V2+V7=0,' }

%%HP: T(3)A(D)F(,);
{ 'V1-V6+V1-V2+V1-V4=0,' 'V2-V7+V2-V1+V2-V3=0,' 'V3-V2+V3-V4+V3=0,' 'V4-V5+V4-V1+V4-V3=0,' 'V5+V5-V6+V5-V4=0,' 'V6-V5+V6-V7+V6-V1=1,' 'V7-V6+V7-V2+V7=0,' }


#28

Nicely done.

When I took first year college physics we called it "Halliday & Resnick", probably the 1st edition. (I must be pretty old because whatever edition it was the section on relativity was a supplement!) I need to dig it out and see if that resistor problem is in there.....


#29

Without nodal analysis, an overkill here, the trick would be redrawing the circuit connecting together the nodes clearly at the same potential, like 2 and 4, for instance.
I should have said "Halliday & Resnick". Actually we called it simply Halliday. Mine was probably the Brazilian 1st edition. I think the resistor cube problem is in the chapter 27 (Circuits).


http://www.amazon.com/Fundamentals-Physics-David-Halliday/dp/0471216437/ref=sr_1_6?s=books&ie=UTF8&qid=1304337310&sr=1-6


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