Recent content by CraigDNeo

Ok, now I think I understand. If below is CAR1 60 foot in 1.083s, 330 foot in 3.112s 660 foot in 4.83s 60 = A*1.17 + B*1.21 + C*1.31 330 = A*9.67 + B*30.08 + C*93.55 660 = A*23.33 + B*112.68 + C*544.23 and I get this in return { a = 63.82661360552823, b = -13.508610964551533, c =...

I put these into the linear solver for a 60 foot in 1.083s, 330 foot in 3.112s 660 foot in 4.83s 60 = A*1.17 + B*1.21 + C*1.31 330 = A*9.67 + B*30.08 + C*93.55 660 = A*23.33 + B*112.68 + C*544.23 and I get this in return { a = 63.82661360552823, b = -13.508610964551533, c =...

so would we do something like this then X = At^3 + Bt2^2 + C*t + D for 0, 60, 330 and 660. Then solve for A, B, C and D. My other question is that when solving for these unknowns, which data point would you start with?

Here is an example for an 1/8th mile (660). Most of the races (90%) are eighth mile. And this is three points of data, and would somewhat less confusing to start with for me at least. Thanks again for your interest and feedback. Car 1 60 feet 1.3389s 330 feet3.909s 660 feet 6.1023s 660 MPH...

And yes we get times at those points too.

Normally we get mph at 660 if the race is only eighth mile. If we are racing quarter mile we get mph at 1320 feet too. Can you help me narrow down the distance with the mph?

the speed of the cars vary allot, but the ones I am working with are moving @115 - 150 MPH through the 660 foot mark in under 6 seconds, and you may come across @140- to over 200 MPH in the quarter mile. These are local track guys, the professionals are doing 1000 foot in under 4 seconds at over...

I think I understand now. Does it change drastically if I want to do the same at the 660, 1000 and 1320 foot markers. Again thank you so much for all the detail you have provided.

So does this change the original posted approach you mentioned: "You have two equations for the first car: Xa=(A+B*ta)*ta^2/2 and Xb=(A+B*tb)*tb^2/2 where Xa=60 feet, ta=1.562 sec, Xb=330 feet, tb=4.335 sec. This is two equations with two unknowns (A and B) and you solve them for A and B."...

Thanks so much for all your help, this is exactly what I needed, I appreciate the feedback and explicit explanations.

my bad, sorry about that.

I redid the break down with the Xa = (A + 3Bta) * ta^2/2 and Xb = (A + 3Btb) * tb^2/2 and got the following Car1 330 = (A + 3B * 4.335) * 9.396 330 = 9.396A + 9.396B * 4.335 * 3 330 = 9.396A + 122.19B (-122.19 from both sides) 9.396A = 330 - 122.19B (divide by 9.396) A = 35.12 -...

Here is my breakdown, please correct me if I am wrong.Car1 330 = (A + B * 4.335) * 9.396 330 = 9.396A + 9.396B * 4.335 330 = 9.396A + 40.73B (-40.73 from both sides) 9.396A = 330 - 40.73B (divide by 9.396) A = 35.12 - 4.33B Now substituting A in the first formula using A = 35.12 - 4.33B I get...

"Multiplication first, then addition. " doh, my bad. Thanks to both of you, let me work the numbers and see what I come up with. I can't thank you enough for all the info you have provided.

Ok, so for Car 1, I have the following: 60 = (A + B * 1.562) * 1.220 330 = (A + B * 4.335) * 9.396 if I use the second formula to solve for A: 330 = (4.335A + 4.335B) * 9.396 35.12 = (4.335A + 4.335B) 35.12 - 4.335B = 4.335A 8.10 - B = A then figured I would use that value for A and plug it...