Thanks Tony !
we went from theory to numbers just to verify the first proposed solution to the problem, with respect to a real event, but identifying the correct physical model is always the main objective of the question I asked in this thread. So thank you for your comments on which I and the...
In fact it shouldn't be negative. In the formula that returns Z0c we have not taken into account h, so the point is a few centimeters above the ground line. In yf=yc+Lsin(θ)−hcos(θ) (as example) the third element of the expression is greater than the sum of the first 2, so already at t0 I get...
Hi,
Setting the spreadsheet I realize that these expressions are a problem, I don't think they are correct: for example Yf gives a negative value, while Yc is positive. I think we should add h to Y0c and for the positions Yr and Yf subtract or add to Yc LSin(θ) without adding the third term of...
Before starting the simulation of phase 2 (free fall), I ask you to confirm the starting data:
##V_0x = V = 2,52 m/s ## (Chord length / 18 frames ...I think it is the most reliable data)
##V_0z = \frac {1}{2}V* \tan(α)##
##ω_0 = \frac {1}{2L}V* \tan(α) - (\frac {3g*L}{h^2+4L^2}*4/24)##...
So, It's my fault in the trascription of the spring rate, due to late night work... Solved!
For vertical suspension rate I read also 500 LB/IN out of the range 0-9 IN.
Please tell me if I have to find others informations...
I try to check if I find the constant .... but I'll check better tomorrow, now I'm about to fall off my chair because I'm sleepy
...nothing at the moment... sorry
SORRY... I had lost a data... there is one more...
VERTICAL SUSPENSION RATE 14 LB/IN (0-9 Inches)
500 LB/IN (< 0 or >9 IN)
SUSPENDED VEICLE MASS 1424 LBs
WHEEL MASS 24 LBs
OK, so 0.68 is an estimate...
If we take the frames as a reference to calculate the time: the entry speed, a few moments before reaching the hole, is 2.61 m/s. After we have 18 frames to run across the arc (2.52 m/s). If we take the ground as a reference to mark the position of the chassis, the...
We have these graphs... but I don't know if that's what we need, I don't know how to interpret them
Is it essential to know how the 4 tractions respond? For the ascent from the dip we didn't ask ourselves...
Sorry, but it's not clear to me... You released a value of 0.68 for J/m, where does...
OK, I'll adapt my mind. But let's remember that in our case the half lengths are different because the centre of gravity is not "in the centre" of the axis.
Sorry but how do you calculate the impulse? Shouldn't it be I/m * Δt ? What is the Δt you used?
I was wondering... since one of the...
Thanks for the question. It's not my intention to hide anything. I also have no more information than anyone searching the public data on this sequence can have. The problem is that I don't have the possibility to recover ALL the data NOW. Last and most importan, what I have raised is a purely...