747 timing

[CSH-HQ]

THANK YOU.  You explainations are very good.
Bruce


>If I'm correct on what you are asking, it's a 3-way, 16-point
>interpolation between the table values.
>
>Think of the point of 53kpa map & 1700 rpm as being someplace inside
>of a box. The four corners of that box, are the four table values
>that surround the defined point. The distance from that point, to each
>of the four walls is estimated, with the final result based on those
>four values.
>
>Works something like this...
>
>Part of main SA table:
>
>map>   50    60
>rpm
>1600   20    24
>1800   22    26
>
>If the map is 53 and the rpm is 1700, first lookup gets the 20 & 24
>degree values, and interpolates between them. The difference between
>these two are divided into 16 points:
>
>24deg - 20deg = 4deg, 4deg / 16points = .25 deg per point
>
>60map - 50map = 10map, 10map / 16points = .625 kpa per point.
>
>53 map @ 1600rpm => 20deg + (3map / .625) * .25 = 21.2 deg.
>
>So, 21.2 deg is the 16 point interp. between 50map & 60map @ 1600 rpm.
>
>(Note: on a 2D table, we would be done, and this value returned).
>(Note: The 3D lookup, just calls the 2D routine three times with
>different values to go between).
>
>The routine then advances a full row to interp. between the
>22 deg and 26 deg values found at 1800rpm.
>
>So, using the same math as above, 23.2 deg is the 16 point interp.
>between 50map & 60map @ 1800 rpm.
>
>
>The third and final interp. is done between the answers of the
>first two lookups. 21.2 deg & 23.2 degrees.
>
>The rpm is divided into 16 points the same. For this example,
>it's right in the middle of table values 1600rpm and 1800rpm,
>with 1700 actual.
>
>So, 21.2 deg & 23.2 deg => 22.2 deg as the actual returned,
>looked up interpolated value.
>
>BobR.
>
>I almost confused myself with this one <g>
>
>
>CSH-HQ wrote:
>> 
>> Some weeks ago, I think RR shared the way a 747 looks up timing.  Would
>> someone mind explain what it does when the MAP value is say 53 K/Pa?.
>> Thanks
>> Bruce
>