Ok, in response to my original post about formulae devoted to determining the importance of DC versus Lap times, I have created/used/modified a mathematical "thing" that can do just that.
I decided to use Matrices (plural of Matrix) as the way of doing this. Matrices are basically groups of numbers that only work as a group, and are great when working with multiple sets of numbers that must all work together. In this case we only have 2 sets of numbers: DC, and Avg lap times. This Matrix can be shown as (DC Avg Lap time)(2 separate numbers). Here is an example: If I had a DC of 7 and a lap time of 1:07.000, or 67 seconds (always convert to seconds), the resulting Matrix will look like this (7 67). We can do this with any set of DC, Avg Lap time, values. Now, every Matrix works by keeping the numbers separate during operations, but we can treat this Matrix as if it were a
Vector (Vectors are numerical representations of Values with both direction, and magnitude) to calculate overall values within the Matrix.
Vectors come in 2 forms: Polar, and Component. Polar are shown like this: (Magnitude, Direction). Magnitude is a number, and Direction is an angle between the X and Y axis (usually), but there can also be a Z axis where needed, but we dont need that). Polar form works by showing how far from the middle of cartesian plane (X = 0, and Y = 0), and at what angle. Component form is shown like this: (X, Y), where X is the X value, and Y is the Y value of a point on a cartesian plane.
We will treat the Matrix as if it is a Vector in component form, where DC = X, and Avg Lap time = Y: (X,Y) = (DC, Avg Lap time), and we will use our previous example: (7, 67). We can find the overall magnitude of the Vector by partially converting it to Polar form. This is how it looks: Magnitude = Square-root-of(X^2 + Y^2), (^2 means squared), where X = DC, and Y = Avg Lap time. Magnitude = Square-root-of(7^2 + 67^2), which equals 67.364... . This is the overall magnitude of the Vector/Matrix: Distance from the middle of the cartesian plane in terms of X an Y.
Now, we can use this to compare other sets of numbers as well, by doing the same with all of the other sets of values. I did however run into a problem with how Avg Lap times affected the outcome of the magnitude, because a slower time would give a bigger number on the Y plane. So, I thought up a form of inverting the times so that the slower time created a lower number, and the lower time: A greater number. This done by using the time distance between one set of values and the leader. We take whatever this number is and double it. The we subtract it from the original time. This means that a slower time will be the same distance from the leader as before, but will end up with a smaller value in the Y plane. Eg: If one time is 67 seconds, and another is 70, the slower one will have 6 subtracted from it ((70 - 67)*2 = 6), and its new time will be 64, which will make its value the same magnitude when compared to 67, but will now work in the right direction. This should be done with all sets of values being compared. Now with this we use Matrices, and Vectors from before to find its magnitude (only when there is also a DC value), and we compare the 2.
I have done this for our last FITT tuning competition to find the placing overall of The Duel class participants that achieved the podium.
GVE R Lap Time Podium
1 Tuner #4 1:10.922 +0.000
2 Tuner #10 1:10.938 +0.016
3 Tuner #7 1:11.045 +0.123
4 Tuner #9 1:11.154 +0.232
5 Tuner #3 1:11.173 +0.251
Drivers Choice Podium
1 Tuner #4 8.8
2 Tuner #7 8.7
3 Tuner #3 8.6
4 Tuner #9 8.6
5 Tuner #6 8.6
Matrices (with inverted Avg Lap times)
Tuner #4 (8.78, 70.922)
Tuner #10 (8.43, 70.906)
Tuner #7 (8.73, 70.799)
Tuner #9 (8.63, 70.69)
Tuner #3 (8.63, 70.671)
Tuner #6 (8.59, 70.639)
Overall placing (magnitude of Matrices)
1 Tuner #4 71.465...
2 Tuner #10 71.405...
3 Tuner #7 71.335...
4 Tuner #9 71.323...
5 Tuner #3 71.195...
6 Tuner #6 71.159...
If this seems not right, the reason is simple: In this scenario, Avg Lap time has a much larger effect than DC, and thus it would be better to aim for a fast time rather than a good DC score. Look at Tuner #10: The tune isnt even in the DC podium, but it is second in the Avg Lap times by only a small margin, thus the DC score had only a small impact on the overall score. If one wished to even out the impact of Avg Lap time compared to DC, you could divide the Avg Lap times by 10, and thus make it less about Avg Lap time and more about DC, but only by a little bit. P.S dividing by 10 to even out DC and Avg Lap times may only work for this instance, Avg Lap time is very close to (DC-1)*10. In other cases it may be more complicated.
This way of finding overall winners by using Matrices and Vectors can be influenced by Multiply/Dividing the DC/Avg Lap times by a given number. This could be used in certain comps to make it more about DC, or more about Avg Lap time, or a simple all around event, where neither value has a larger effect.
This is all I can think to do, and I believe I have covered everything. If ANYONE does not understand something: please tell me, and I will elaborate. Also, in any upcoming FITT events that plan on using this system: please inform me so I can help with organisation.
Sound good?