Maximum Radius for Wall of Death

[mfb]

I used $g \approx 10 ms^{-2}$. The difference is just 2 %.

 

[Orodruin]

Just as a side note, $a = v^2/r$ is just the centripetal acceleration. If you want to find the total acceleration experienced by the rider, use Pythagoras' theorem to find that
$$
a_{\rm tot}^2 = g^2 + \frac{v^4}{r^2}.
$$
For a total acceleration of $Ng$ this would give a velocity
$$
v = \sqrt[4]{N^2 - 1}\, \sqrt{gr},
$$
indicating that for a fixed target acceleration, $v$ needs to grow as $\sqrt{r}$ (this was true already for the centripetal acceleration and follows directly from basic dimensional analysis). The angle is always the same for a given acceleration, $\cos(\theta) = 1/N$, where $\theta$ is the angle of the rider from the vertical position.

 

[Jimbo54321]

I am the first to admit that I am not good at mathematics and have struggled all my life, but I do try, and I will persevere in improving my understanding.
I can see from other threads that this forum is used by Ladies and Gentleman that are so far above me in the league tables, that when it comes to understanding mathematics, physics and science, you guys leave me feeling like a Bum at a banquet. LOL.

Anyway, the only reason I questioned my math was to answer a nagging thought of how many G's Patch would be pulling at 48mph (as stated in the program), but when I worked out that there is no way he could do 48mph, I had to question (1) my math, or (2) the programs statement.

I'd like to thank you guys for helping me, mfb, Doc Al and Orodruin you three have given me a lot of self homework, (trying to decipher the formulas you used) PMSL.
But I will strive to understand this math. Oh the late nights.
Thanks again.

 

[Orodruin]

What is to be said on the topic has been said and several posts have already been deleted. It is time to close this thread.