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I am studying the sequence and data in your file and something look curious, after each 115 ID, the data in D8 is different even if D1 to D7 is the same. There is clearly a link with the 0-1-2-3 in D8 of 115H
Do you know if D8 from PCM is a checksum or data ?
RD119,783217D0000A009
RD119,753217D0000A0018
RD119,763217D0000A0027
RD119,773217D0000A0036

I found another thing in the sequence, the ID 231 and 261 come together and only after the ECM transmit the 115H in the sequence with 2 in D8
 

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Discussion Starter #24
some look clear the data in 17D -111 - 141 -169 and 179 depend of the sequence

I made a Excel file with color, the sequence is more clear in that
I hope it will work. try rename the file for xlsx
Unfortunately I only have Excel 2003 with add on packs.
It could not cope with all the cell formatting in your newer version. :(
Maybe Google sheets?

Here is a slightly different angle with a file captured using the Saleae logic probe and it's CAN analyser.

This is the ECM and MCM talking via the CAN bus from IGN turn on.
It give a better visual idea of timings etc but you need to install the free Saleae software to view it.

Logic analyzer software from Saleae

Both units take slightly different amounts of time to come on line and start transmitting after the CAN wakes up.

We get some preamble and error packets with no acknowledgements as one unit starts transmitting before the other.
But after about 140ms we get into the routine 10ms interleaved packet transmission that continues ad infinitum.

So just as a reminder...

The ECM (ECU) sends out the 115, 17D, 24F, 318 packets
The MCM (IMA) sends out the 111, 141, 169, 179, 231, 261, 307 packets.

Rename the attached file extension to ".logicdata"
 

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Tanks for the reminder I had a bad point of view for the 17D transmitter
I continue to work on 17D D8. I think it is a checksum but didn't find the solution for now

I learned about google document and I hope my link will work
 

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I willl need nore data for the 17D , I extracted all diferent possibilities of your file and that what I found
when We have the same data for at D1 to D7, the D8 is the lower when it start by 3
when it start by 2 it is +1
when it start by 1 it is +2
when it start by 0 it is +3
there is a scheme is that but i did not find a equation at this moment. Maybe with more data it will show something more clearly.
can you send me a file like the first one (txt file) with lot ot 17D and if you can catch data with a lot of 0 it is more easy.
for the $115 is was the fact that there was a place with only one column with number 1 and other with 2 that help me a lot

D1D2D3D4D5D6D7D8
9,903217D0000A009
11,783117D003603A00D
9,983217D003403A00F
9,953217D00403A0011
9,913217D0000A0018
43,031117D003603A001C
11,713117D003403A001E
9,923217D0000A0027
43,041117D003603A002B
11,723117D003403A002D
9,933217D0000A0036
43,011217D003603A003A
11,733117D003403A003C
 

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Discussion Starter #27
Don't worry about 17D for now.
I can't control that data of collect simpler valid packets for analysis.

Packet 115 is the only one I will be altering to start with.
 

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I just font the solution for the 17D
I hope later I will help

For the one with D8 start by 3 : add each half byte then reverse the last 4 bit and add 1

17D 0 0 36 0 3A 0 0 3A

0 + 0 + 3 + 6 + 0 + 3 + A + 0 + 0 = 16 (hex) or 0001 0110 (bin)

Reverse all bit of 6 or 0110 = 1001

Add 1 = 1010 = the A of 3A

--------------------------------------------------------------------------

For the one with D8 start by 2 : add each half byte then reverse the last 4 bit and add 2

17D 0 0 36 0 3A 0 0 2B

0 + 0 + 3 + 6 + 0 + 3 + A + 0 + 0 = 16 (hex) or 0001 0110 (bin)

Reverse all bit of 6 or 0110 = 1001

Add 2 = 1011 = the B of 2B

--------------------------------------------------------------------------

For the one with D8 start by 1 : add each half byte then reverse the last 4 bit and add 3

17D 0 0 36 0 3A 0 0 1C

0 + 0 + 3 + 6 + 0 + 3 + A + 0 + 0 = 16 (hex) or 0001 0110 (bin)

Reverse all bit of 6 or 0110 = 1001

Add 3 = 1100 = the C of 1C

--------------------------------------------------------------------------

For the one with D8 start by 0 : add each half byte then reverse the last 4 bit and add 4

17D 0 0 36 0 3A 0 0 0D

0 + 0 + 3 + 6 + 0 + 3 + A + 0 + 0 = 16 (hex) or 0001 0110 (bin)

Reverse all bit of 6 or 0110 = 1001

Add 4 = 1101 = the D of 0D
 

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I feel like a mouse stuck in his wheel.
I just see a pattern in the ID 111 - 141 - 169 - 179 I am sure they have checksum similar
For the 24F I think there is no checksum but not certain.
 

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I just font the solution for the 24F
there is a checksum and the calcul is the same as the 17D
For the one with D3 start by 3 : add each half byte then reverse the last 4 bit and add 1

24F 78 20 3F

7 + 8 + 2 + 0 = 11 (hex) or 0001 0001(bin)

Reverse all bit of 1 or 0001 = 1110

Add 1 = 1111 = the F of 3F
--------------------------------------------------------------------------
For the one with D3 start by 2 : add each half byte then reverse the last 4 bit and add 2
For the one with D3 start by 1 : add each half byte then reverse the last 4 bit and add 3
For the one with D3 start by 0 : add each half byte then reverse the last 4 bit and add 4

just dont forget to keep only the 4 leat significant bit of the result
so we have now 17D and 24F with the same calcul. I hope others will be the same 😀
 
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