My first thought after (finally) getting a 39gII was that it was really worrying as a basis for the HP Prime, a new flagship for HP.

I intended to post some issues I had, but I had opportunity to run a quick calculation and compared my (emulated) 48gx (though my 50g would have been similar) to the 39gII. While involving a unit calculation is probably extremely unfair (I can't imagine improving the 50g model much), it does hilight other issues I have along the way. I would welcome any suggestions on improving the use/flow of the 39gII as well.

So the (lunch) problem was to figure out if the first atomic bomb, apparently based around 100 lbs of uranium, was terribly inefficient given the quoted fact that no six inch pure uranium nuggets are floating around in space, since that size would go critical (as opposed to the potential for other pure nuggets). (These figures aren't exact, but that was the remembered facts.)

A quick check of The Elements says the density of Uranium is 19.05 g/cm^3. A quick Google says the volume of a sphere is 4/3*pi*r^3.

On the 48gx:

100 rShift Units Mass LBgives 2.381E-3_m^3 as the volume of 100 lbs of Uranium. Switching modes to fix 6:

19.05 g rShift Units VOL rShift cm^3

/ lShift Units UBASE

rShift MODES CHOOSE dwn OK right 6 OK OK

yields 0.002381_m^3. Compute the diameter with:

3 * 4 / lShift pi / lShift >Num 3 rShift x_root_y 2 *

so 0.165675_m in diameter

rShift UNITS LENG lShift IN

or 6.522644_in in diameter. So not too wasteful.

Contrast with doing this on a 39gII:

Turn it on and wait the two seconds to boot up.

100 Math Units dwn*7 right dwn*3 OK / 19.05 Math right dwn OK(Couldn't figure out best way to enter 19.05_g/cm^3.)

* 1 Math up*3 right dwn OK ENTER

Giving us 5.249_(lb*g^-1*cm^3) which we base with:

Math up*4 right dwn*4 OK shift ANS ENTER

which gives us 2.381E-3_m^3 then we change modes to Fix 6:

shift Modes dwn Choose dwn OK right 6 OK HOME

Since the display doesn't change the old answers (!) we do:

shift ANS ENTER

and see 0.00238106_m^3 (what happened to fix 6???) of uranium. Then we compute the diameter of a sphere of that volume:

2 * ( 3 shift NTHROOT ( 3 / 4 * up COPY bs*4 / shift pi ENTER

(We lost precision, but we can't take NTHROOT of a unit and I couldn't figure out a way to strip units from ANS.)

So now we have 0.165675 meters which we convert to inches

Math right dwn*2 OK up COPY Math dwn*2 right OK , 1 Math right dwn*15 OK ENTER

And we have 6.522638_inch.

I think you can see some annoyances :)