Nice presentation DMW.

As you also state in your document, the problem with the Z32 – assuming homophonic substitution – is that because of its freakish multiplicity it is impossible to verify any solution without "extra information" of what letters the symbols are supposed to be.

So that means the Zodiac was ALA and that he considered himself a bomb?

As you also state in your document, the problem with the Z32 – assuming homophonic substitution – is that because of its freakish multiplicity it is impossible to verify any solution without "extra information" of what letters the symbols are supposed to be.

Agreed.

Although the Z32 is unique in that there IS extra information — from the map and the postscript.

I originally created this solution because I got fed up reading other Z32 solutions that either ignored major pieces of information, misunderstood polar coordinates, or leaned on obscure numerological systems that aren’t even hinted at anywhere else in Z’s communications. I though that it must be possible to for there to be a solution that avoided all these problems, and rather than sit there complaining about it, I decided to create a solution like that.

My mind got blown, a little, when I discovered that ALA’s house was exactly at the 10:00 position. I hadn’t expected that, and to me it elevates this solution from just an intellectual exercise to one that has a (small) chance of being what the killer intended.

But I agree with you completely — this is not a decisive solution, and it is doubtful that one is even possible.

In 2012, I came up with this solution:

The cipher has multiple solutions, it is just too short. I doubt, however, that it leads to ALA’s house (didn’t he live in a trailer anyway?).

Additional read with long/lat calculation:

viewtopic.php?f=97&t=802&p=6128&hilit=Mount+Diablo+Cipher+pdf#p6128

QT

In 2012, I came up with this solution:

The cipher has multiple solutions, it is just too short. I doubt, however, that it leads to ALA’s house (didn’t he live in a trailer anyway?).

QT

I LOVE this! We are thinking alike (though you beat me to it by a few years).

Just FYI — ALA lived with his parents at 32 Fresno St., but he also had a couple of trailers he stayed in at other locations. (I don’t remember how many trailers.)

QT, why is that document named "Folie 1"? It means "Madness 1" in French! ^^

I wonder if Z really expected people to find a solution to all this.

@Claypooles: Folie is German and means foil, sheet. In french madness..that’s what it actually is (you should know, you have Macron)

@DMW: Yes the cipher is short, thus can deal with a few words only. Z gave the hint with inches, radians. Still multiple solutions possible. Don’t worry about the years, the exact GPS data of Z’s residence would beat all of it ;D

QT

The following is the calculation of e.g. ‘2.2 inch’ and ’18 radians’ (metric, degrees clockwise):

Circumference of the earth
Diameter x PI
12.742 x 3.14159 = 40,030.14 km

Latitude/Longitude Mount Diablo
Latitude: 37.881666666667
Longitude: -121.91277777778
Degrees: 37° 52′ 54” North , 121° 54′ 46” West

Degrees
1 arc degree = 60 arc minutes = 3600 arc seconds

Degrees of latitude
1 arc minute = 1 nautic mile = 1.852 km
1 arc degree = 1.852km x 60 = 111.12 km
1 arc second = 1/60th nautic mile = 30.87m

Degrees of longitude
1 arc degree[lon] = cos(arc[lat]) x 1 arc degree [lat]
1 arc degree[lon] = cos(37.881667) x 111.12 km = 87.7049 km
1 arc second[lon] = 24.36m

Inches
1 inch = 6.4 miles (‘one inch equals approximately 6.4 miles‘ [philips map])
2.2 Inch = 14.08 miles = 22.66 km

Degree vs. Radians
PI x RAD = 180°
>> Degree = RAD x 180° / PI
18 Rad = 1.8 RAD (Radians always in steps of ten)
Degree = 1.8 x 180° / 3.14159 = 103.13249°(East-South-East)

Magnetic declination
1970-06-26: 16.8799° East
Result: 103.13249° + 16.8799° = 120.0124° (geographic)

Pythagoras
We use the 90.0° as ‘a’. Thus:
alpha = 120.0124° – 90.0° = 30.0124° (East)
c = 14.08 miles
a = cos[alpha] x length[c] = 0.8659 x 14.08 miles = 12.1921 miles
b = root(14.08^2 – 12.1921^2) = 7.0427 miles
"From Mt. Diablo go 12.1921 miles East, then 7.0427 miles South"

Position change
Longitude change: 12.1921 miles x 1852m / 24.36m = 926.9199 arc sec = 15′ 26.9199” EAST
Latitude change: 7.0427 miles x 1852m / 30.87m = 422.5164 arc sec = 7′ 2.5164” SOUTH

Position new
37° 45′ 51.4836” North , 121° 39′ 19.0801” West
latitude 37.764301 longitude 121.65530002777778

Hope it’s correct. However I end up at some landfill. This time I did the calculation with the windrose clockwise, which was vice versa last time (ending up in Lafayette). Just to show the method.

QT

QT — although I really like your solution, I do not think this is the best way to make these calculations.

The calculations should be made the same way that Z did it — that way you will get the same results he did. And I doubt he did all that.

I think he simply measured on the map with a ruler and a protractor. It’s much easier. The result will be less precise, but I don’t think Z was so concerned about that.

I suggest you try this — it might move the result slightly, and take you out of that landfill.

Well, even if Z had used the ruler, he will end up at the same spot (with that ‘solution’). It could be an indication that the solution is not correct, although it fits into the cipher structure.

The calculation, btw, is simple navigation like Captain Cook might have used it. And Z had some ‘military’ skills: Mag North, 0-3-6-9 instead of N-E-S-W as directions, cryptology, shooting..not to forget about the Wingwalkers. Thus, he could have been educated in navigational skills, too.

Depending on what type of windrose (clockwise / non-clockwise) Z had used, I end up with the following two locations (based on 2.2 inch, radian 18):

Clockwise (West)
Lat 37.8970047, Lon -122.2094614
Some empty field near Orinda (370, Camino Pablo…the old St. Stephens Drive address imo was wrong due to a trigonometric calculation error..srry)

Non-clockwise (East)
Lat 37.76429003, Lon -121.6553277
Slightly North of Altamont Pass Road

As well as based on 1.1 inch, radian 18 (second cipher ‘solution’):

Clockwise (West)
Lat 37.88933585, Lon -122.061121
Some ‘promising’ address in Walnut Creek: 105, Arlene Drive

Non-clockwise (East)
Lat 37.82297852, Lon -121.7840539
Some forest location near Whipsnake trail

Thus, if you draw a 1.1 inch line towards 1.8 radians, non-clockwise, considering a 16.7899° declination (Mag. North) to the East, you end up at e.g. 105, Arlene Drive. If you believe in a solution of 2.2 inch (see presentation), it’ll be 370, Camino Pablo (gps, not google maps address), opposite Wagner Ranch Elementary School. If Z used the windrose clockwise, either Whipsnake trail (1.1 inch) or Altamond Pass Road (2.2 inch).

All of the previous is actually wrong if a different cipher solution is meant to be used..but if the cipher solution was correct, the addresses above are valid, imo.

QT

Well, even if Z had used the ruler, he will end up at the same spot (with that ‘solution’).

QT

I don’t agree. This would only be true if the map was perfect. I doubt that it is. I DO think you would end up at a spot close to the one you calculated, but I do not think it would be exactly the same.

Then it’s not a map ;D true map could be not precise but result is sort of the same (red 2.2 inch, green 1.1 inch..the Eastern 2.2 inch spot is not even on the map). Lines to the left is if Z had drawn the degrees non-clockwise (West), lines to the right if he draw them clockwise (East). Yes, they have different angles compared to West/East due to ~16% declinatin (magnetic North).

Advantage of calculation is no more than getting the exact gps address instead of the estimated end of the 2.2 / 1.1 inch line (except map ‘inaccuracies’ it is the same location..).

QT