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Solved ✔ vz commodore odometer algorithm
20-02-2023, 14:07 PM,
Post: #1
KHolden  vz commodore odometer algorithm
Hello! I can calculate the original mileage from the dump attached, however what I’m really wanting to know is how to reapply an exact odometer correction.

Note: The dump attached is one I found online since I’m using a bench cluster.

For example, the algorithm has two components that are skewed across 20 different numbers. So like, 12x 8919 and 8x 8920…my question is this:

How do we know which amount of both to put in when we’re calculating the replacement reading?

Is this a well-known algorithm? For example, the cluster manufacturer was Australian Arrow Pty. Ltd. (AAPL), Australia’s division of Yazaki.


Attached Files
VZCluster.zip
File Type: .zip
Downloaded: 4 times
Size: 520 bytes

20-02-2023, 14:54 PM,
Post: #2
RE: vz commodore odometer algorithm
Pretty simple algo, that file has 241095 Km.

Modified Yazaki, without the complement words.

2f17 X 20 = 241100
Minus 5 for the 5 2f16 = 241095

If someone took the time to help you, the least you can do is take a few seconds and press two buttons! 

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20-02-2023, 14:59 PM,
Post: #3
RE: vz commodore odometer algorithm
Thankyou very much for the response smile

I’m very new to this; what is a complement word?

Also, how do they get 5x 2f16 and not, say, 7x? That’s the bit that’s confusing me as I try and reverse the algorithm’s direction to encode a reading instead of decoding it.
20-02-2023, 15:15 PM,
Post: #4
RE: vz commodore odometer algorithm
7x 2f16 would make it 241093

The hard way, convert each word to decimal, then add them together.

To get a new value, mileage / 20 to hex. To make it simple, use multiples of 20.

If someone took the time to help you, the least you can do is take a few seconds and press two buttons! 

Thanks [Image: tyl_add.png] + [Image: postbit_reputation.gif] reputation.
20-02-2023, 17:01 PM,
Post: #5
RE: vz commodore odometer algorithm
Okay, I’ve figured it out smile this is what I did:

The first part is calculated by dividing the reading by 20 and rounding it up.

The second part is calculated by doing the same, but then rounding it down.

The fractional component that remains after you divide the reading by 20 is multiplied by 20 to give you the number of times the second component should be repeated.

Obviously then, the number of times the first component appears is determined by 20 minus the number of times the second one appears.


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