Little explorations with HP calculators (no Prime)

[pier4r]

good point!

[Zaphod]

rprosperi Wrote:My suggestion is to leave the miniUSB cables plugged-in to both calculators, and only plug/unplug the USB-A (PC side) connector as needed. The USB-A connectors are much more robust than the small and delicate mini-USB.

Apparently not....., the opposite in fact

USB-A has a mere 1500 mating cycle rating

See section 5.3 of this current example by molex
https://www.molex.com/pdm_docs/ps/PS-48037-001-001.pdf
(same story with the A socket)
https://www.molex.com/pdm_docs/ps/TS-48037-001-001.pdf

Mini-B is 5000 mating cycle rated

See section 4.2.2 of this active current example by molex:
https://www.molex.com/pdm_docs/ps/PS-51387-003-001.pdf

and Micro USB socket is 10,000 mating cycle durable

See section 3.3.2 on this active current example by molex again:
https://www.molex.com/pdm_docs/ps/PS-47589-001-001.pdf

Goes against what you intuitively think based on the size of it, but there it is.

[rprosperi]

(04-02-2018 02:24 PM)Zaphod Wrote:  Goes against what you intuitively think based on the size of it, but there it is.

Interesting theory.... my real-world experience is different. I have been asked to replaced many device-side connectors, but rarely a PC-side. No doubt this is due to more careless handling/operation, but whatever the reason, these simply fail more often.

But more importantly, if the USB-A socket on the PC fails, it is easily replaced (even if needed, most PCs have more than needed), and cost is about $10 for a new board, or cheap hub.

If the mini-USB port of the 50g is damaged, you must replace the 50g.

[Zaphod]

(04-02-2018 05:34 PM)rprosperi Wrote:  I have been asked to replaced many device-side connectors, but rarely a PC-side. No doubt this is due to more careless handling/operation, but whatever the reason, these simply fail more often.

It's not exactly a theory , these are manufacturers specs.

I would suggest failure is pretty much always prematurely down to careless handling/accidents, but the micro/mini have a better chance of serious board damage... for it to actually 'wear out' as per the data-sheet is virtually unheard of.
Two distinct kinds of failure.

The plug will act as a lever when an external 'accident' happens and rips the tracks off (if you're very very lucky and the board is tough enough, the lead-free soldering breaks free before the actual tracks, otherwise an accident will just rip the tracks to bits, often destroying any chance of repair.)
They rely on two soldered lands on the board for physical security (as well as a small amount offered by the connection tracks.)

USB A sockets are a bit more substantial and sometimes (but not always) have 'through-hole soldered' lugs for physical security. micro and mini sockets just aren't big enough for this kind of help.

For actual plugging in and unplugging if you're gentle doing so, the micro and mini stand a better chance of wear resistance.

Quote:If the mini-USB port of the 50g is damaged, you must replace the 50g.

Particular reason ?

If the socket itself is damaged and still secure on the board , surely a replace of the socket is on the cards (with the appropriate smd rework kit) ?

[rprosperi]

(04-02-2018 06:17 PM)Zaphod Wrote:  

Quote:If the mini-USB port of the 50g is damaged, you must replace the 50g.

Particular reason ?

If the socket itself is damaged and still secure on the board , surely a replace of the socket is on the cards (with the appropriate smd rework kit) ?

I'm really making a general statement for that set of the audience that aren't skilled enough to do their own SMD-level repairs (maybe 98%). More than likely, these folks are careful enough to not damage stuff anyhow, though of course accidents can happen to anyone.

Though the specs you cite are indeed interesting (and somewhat surprising to me) I stand with my earlier advice. Real world behavior has proven that you're more likely to damage the 50g than the PC if you constantly are connecting/disconnecting the device.

While an argument could be made that by leaving the miniUSB connected to the 50g all the time, you are in fact more likely to cause damage as a fall landing on the top of the machine with cable inserted could break the connector or its interface to the board more easily than if it fell alone, such contrived conditions are outside the 'normal' use scenario, IMHO.

[pier4r]

(03-17-2018 08:55 PM)pier4r Wrote:  A question still remains.
Question to the 71B masters. Can basic on the 71B receive a program (or a reference to a program) as an argument? Or can do indirect addressing?

A kind user solved this problem for me in PM.

Code:


Yes, the 71B can call a subprogram in this way:
10 A$="MYSUB"
20 CALL A$(3)
30 END
40 SUB MYSUB(X)
50 PRINT X
60 END SUB

edit: and while I thank Zaphod for the manufacturer datasheet, I prefer to destroy 10 PC hubs than one 50g. No I am not skilled enough to fix the board components.

[John Keith]

I seldom use the USB port for data transfer, mostly the SD card. OTOH, I'm not sure how rugged the SD card slot is either.

[Zaphod]

rprosperi Wrote:I'm really making a general statement for that set of the audience that aren't skilled enough to do their own SMD-level repairs (maybe 98%). More than likely, these folks are careful enough to not damage stuff anyhow, though of course accidents can happen to anyone.

Though the specs you cite are indeed interesting (and somewhat surprising to me) I stand with my earlier advice. Real world behavior has proven that you're more likely to damage the 50g than the PC if you constantly are connecting/disconnecting the device.

Ah right , gotcha , yes , even I’m careful of products when a micro usb is plugged in , because it’s ‘prone’ ... a potential accident waiting to happen... I try and place the cable on a hard surface with little chance of the cable imparting any force on the socketry.
I do this sort of repair work as a day job, so whilst I can do the repair, it’s something I’d rather avoid causing in the first place .... but I don’t have any qualms about actually plugging in and out per se.

Quote:I'm not sure how rugged the SD card slot is either.

Hmmmm, there’s another problem socket (not particularly on the 50), often gives me headaches. Lol

[pier4r]

For the ones going to paste their text with rpl and digraph in emu48 the following link may be interesting, because otherwise one has errors and may be puzzled.

https://groups.google.com/forum/#!topic/...kKpV03RRk4

In short the digraph have to be converted to go in the emulator.
There is also a program reported in this page http://hp.giesselink.com/emu48.htm and Tim W. suggested also that would be possible to connect the conn4x (that includes the ability to convert text to hp obects) to emu48 using a virtual COM port. For example this: http://com0com.sourceforge.net/ . Another possibility is using DDE, but with that I am not sure the text is converted to an object.

PS: although I know that emulators are a great work, I normally try to use the real 50g. Only when I have to transfer back and forth a lot of data then I have to admit that emulators / desktop app are more handy.

Update: the ascii to bin conversion program did not work. Likely due to my encoding settings in notepad++ or notepad2-mod. I used the com0com null modem, set the emulator using the COM7 and the conn4x works, seeing the calculator on COM7. In this way it is relatively easier to pass text to the emulator. It is a bit less faster than before, but still faster than a real 50g, if the data to pass is a lot.

[RMollov]

(03-27-2017 07:39 PM)Gerson W. Barbosa Wrote:  

(03-27-2017 12:14 PM)pier4r Wrote:  

Let's divide one of the right-triangles into four right-triangles (two with sides lengths equal to r and x and two with side lengths r and 1-x) and a square with side r. Then it's area is

S = r*x + r(1-x) + r^2

S = r^2 + r

The area of the larger square, witch we know is equal to 1, is the sum of the areas of these four right-triangles plus the area of the smaller square in the center, with side 2*r:

S = 4(r^2 + r) + (2*r)^2 = 1

Or

8*r^2 + 4*r - 1 = 0

Here I am tempted to just take my wp34s and do 8 ENTER 4 ENTER 1 +/- SLVQ and get a valid numerical answer for the quadratic equation, but I decide to go through a few more steps by hand and get the exact answer:

r = (sqrt(3) - 1)/4

Would you show those steps here? I stopped before them thinking that's enough...

Thanks,

*** edit: I mean only if you derived this solution directly from the geometry and not by playing with the equation WolframAlpha style

[pier4r]

Some notes for those that are in my previous situation and for myself in the future (when I will forget).

1. simple but effective backups on the SD card (50g).
Of course one can make or search proper backup programs that do everything automatically but in the worst case do:
(a) tidy up the SD card main folder with the help of libraries like the sdfiler (see http://www.hpmuseum.org/forum/thread-10644.html )
(b) select all the things in port0 (HOME) to be copied with the built in filer (the same can be done with port1 and port2 files)
(c) copy them on the SD card
(d) with the SDfiler tidy up again the sd card, for example moving the new files to a proper folder.
(e) repeat as wished but be careful, the sdfiler sometimes doesn't really see many files in a folder

2. enjoy the power of conn4x (connectivity kit software between PC and 50g plus usb drivers for the 50g). This will require winXP (mostly). Wait a moment, it means that you should keep a winXP functioning device as well for your hp 50g! Or you play around with virtual machines.

conn4x can quickly transfer data in textual form. Use the "edit as text file". It is really flexible. You can create new folders, rename them, and then edit the content (converting them to programs or so) with ease.

Also you can discover how folder works, seeing the content (as text) of one filled folder on the 50g. Alternatively go search threads like this: http://www.hpmuseum.org/forum/thread-8054.html with the explanations of DavidM.

also conn4x is useful to pass a lot of data in text format (plus digraphs) to the virtual calculator as well. The virtual calculator doesn't swallow easily digraphs or line endings.

[grsbanks]

(05-21-2018 12:53 PM)pier4r Wrote:  2. enjoy the power of conn4x (connectivity kit software between PC and 50g plus usb drivers for the 50g). This will require winXP (mostly). Wait a moment, it means that you should keep a winXP functioning device as well for your hp 50g! Or you play around with virtual machines.

The 50g connectivity kit works fine on any version of Windows as long as you remember to deactivate driver signing enforcement on versions of Windows that support this when you install the USB driver.

[rprosperi]

(05-21-2018 01:04 PM)grsbanks Wrote:  The 50g connectivity kit works fine on any version of Windows as long as you remember to deactivate driver signing enforcement on versions of Windows that support this when you install the USB driver.

1 +

I've used conn4x for years under Win-7 Pro x64 with no issues at all with a 50g.

With a 48GX, I can't seem to get conn4x to capture screen shots, but otherwise no issues with 48GX either.

[Gerson W. Barbosa]

(05-06-2018 06:13 AM)RMollov Wrote:  

(03-27-2017 07:39 PM)Gerson W. Barbosa Wrote:  Let's divide one of the right-triangles into four right-triangles (two with sides lengths equal to r and x and two with side lengths r and 1-x) and a square with side r. Then it's area is

S = r*x + r(1-x) + r^2

S = r^2 + r

The area of the larger square, witch we know is equal to 1, is the sum of the areas of these four right-triangles plus the area of the smaller square in the center, with side 2*r:

S = 4(r^2 + r) + (2*r)^2 = 1

Or

8*r^2 + 4*r - 1 = 0

Here I am tempted to just take my wp34s and do 8 ENTER 4 ENTER 1 +/- SLVQ and get a valid numerical answer for the quadratic equation, but I decide to go through a few more steps by hand and get the exact answer:

r = (sqrt(3) - 1)/4

Would you show those steps here? I stopped before them thinking that's enough...

Sorry for the late reply! I stopped following this thread since page 8 or 9, so I’ve completely missed your request.
The next steps are simply the manual solution of the second-degree equation using Baskhara’s formula:

r = (-4 + sqrt(4^2 - 4*8*(-1)))/(2*8)
...
r = (-4 + sqrt(48))/16
...
r = (sqrt(3) - 1)/4

Pier has provided a drawing that illustrates the solution.

Gerson.

[RMollov]

(05-22-2018 03:03 AM)Gerson W. Barbosa Wrote:  

(05-06-2018 06:13 AM)RMollov Wrote:  Would you show those steps here? I stopped before them thinking that's enough...

Sorry for the late reply! I stopped following this thread since page 8 or 9, so I’ve completely missed your request.
The next steps are simply the manual solution of the second-degree equation using Baskhara’s formula:

r = (-4 + sqrt(4^2 - 4*8*(-1)))/(2*8)
...
r = (-4 + sqrt(48))/16
...
r = (sqrt(3) - 1)/4

Pier has provided a drawing that illustrates the solution.

Gerson.

Thanks, Gerson
As I already mentioned, I'd be interested if that could be derived directly from geometry. Now I'm happy that's not the case.

[pier4r]

The following problem can be used to explore list functions. For example on the 50g. Useful list libraries that expands the built in ones: listExt ( http://www.hpmuseum.org/forum/thread-8555.html ) , goferlist ( https://www.hpcalc.org/details/6529 ) and lsort ( https://www.hpcalc.org/details/2828 ).


The problem: there are 100 blue Christmas balls and 40 white Christmas balls in a line, randomly ordered. There are Anna and Berta that needs balls for their Christmas trees.

They decide that Anna will pick a continuous chunk of the line of length 70, for example between the position 3 and the position 72, the rest will be picked by Berta.

Could they find a way to ensure that Anna and Berta pick both 50 blue balls and 20 white balls? If yes, is a special arrangement of the balls needed to achieve it?

[ijabbott]

(12-26-2018 09:34 PM)pier4r Wrote:  The problem: there are 100 blue Christmas balls and 40 white Christmas balls in a line, randomly ordered. There are Anna and Berta that needs balls for their Christmas trees.

They decide that Anna will pick a continuous chunk of the line of length 70, for example between the position 3 and the position 72, the rest will be picked by Berta.

Could they find a way to ensure that Anna and Berta pick both 50 blue balls and 20 white balls? If yes, is a special arrangement of the balls needed to achieve it?

There will always be a contiguous segment of 70 balls containing exactly 50 blue balls and 20 white balls. If there is a starting position that produces more than 50 blue balls, then to even things out there must be some other starting position that produces fewer than 50 blue balls. Since shifting the starting position by one place changes the number of blue balls by at most 1, there must be some starting position between those two positions that produces exactly 50 blue balls.

(This uses a discrete variant of the intermediate value theorem.)

[DavidM]

To explore this, I put together a couple of helper apps.

The first simply creates a randomized list of 1s (blue balls) and 0s (white balls):

Code:

Balls
\<<
   @ create a list containing 100 blue
   @ and 40 white balls
   @ (1 = blue, 0 = white)
   1. 100. LMRPT
   0. 40. LMRPT
   +

   @ randomize the list
   LSHUF
\>>

"Balls" creates a randomized list of 100 1s and 40 0s in about 0.1s.

To see how many contiguous runs of 50 blue/20 white "balls" there were in a list, I created this routine:

Code:

AnnaBerta
\<<
   @ result list
   { }

   @ check all contiguous groups of 70
   1. 71. FOR x

      @ extract 70 balls starting from position x
      OVER x DUP 69. + SUB

      @ count the white balls
      0. LCNT

      @ if 20, add x to result list
      20. SAME
      { x + }
      IFT
   NEXT
\>>

"AnnaBerta" takes the list created by "Balls" and checks each contiguous subgroup of 70 elements for the proper mixture of blue/white elements. I assumed that no wrap-around is appropriate for contiguous groups, so I only check starting positions of 1-71.

The smallest result I recall seeing was 1, ie. a single contiguous group in the target list. The largest would be 71, since it's possible to come up with an ordered list where all contiguous subgroups meet the criteria. I've yet to see that in random testing, though.

Execution time of "AnnaBerta" varies depending on the number of found starting points. A list with only 1 "match" completes in about 2.4s, and the opposite extreme (71 matches) completes in about 3.1s.

[DavidM]

After running a bunch of iterations of the previously-posted routines, I became curious about the distribution of counts of matches (where a "match" is defined by a proper mix of 50/20 colored balls).

Knowing that it would be too slow to attempt this on a 50g, I opted to attempt this with python running on my laptop (I'm still in a learning mode with python -- please be kind).

I came up with the following to generate one million random lists of 100 blue/40 white balls, then summed the individual counts to see how they were distributed:

Code:

import random

def random_list():
    result = [1]*100 + [0]*40
    random.shuffle(result)
    return result

def match_count(thelist):
    matches = 0
    for i in range(71):
        sublist = thelist[i:i+70]
        if sum(sublist) == 50:
            matches += 1
    return matches

def main():
    result = {}
    for i in range(1,72):
        result[i] = 0
    for __ in range(1000000):
        thelist = random_list()
        matches = match_count(thelist)
        result[matches] += 1
    for i in range(1,72):
        print('%d\t%d' % (i,result[i]))

if __name__ == '__main__':
    main()

The final result of the above is a tab-delimited list of counts and quantities, and the following is a sample of one run:

Code:

1    60726
2    60743
3    60133
4    59444
5    58566
6    57093
7    55206
8    53189
9    51518
10    48693
11    45882
12    43496
13    40126
14    37631
15    34314
16    31327
17    28141
18    25558
19    22785
20    19674
21    17479
22    15290
23    13129
24    11113
25    9502
26    7924
27    6637
28    5407
29    4362
30    3518
31    2719
32    2233
33    1763
34    1291
35    950
36    690
37    509
38    399
39    282
40    177
41    123
42    100
43    54
44    40
45    24
46    25
47    5
48    3
49    4
50    1
51    2
52    0
53    0
54    0
55    0
56    0
57    0
58    0
59    0
60    0
61    0
62    0
63    0
64    0
65    0
66    0
67    0
68    0
69    0
70    0
71    0

Placing that data into an Excel spreadsheet and plotting the result gives a very familiar shape:



I suspect the probability of randomly generating a list that has 71 "matches" is quite small.

[pier4r]

David I am not sure I understood your test. I am not sure I decoded the data properly.

For example:

Code:


1    60726
2    60743
3    60133
4    59444
5    58566
6    57093
7    55206

It means that out of 1 million attempts, 60726 of them had at least one list of 70 elements that matches the requirement? 60743 times there were two matching lists?