Model 3 crash @ +100MPH

[DopeGhoti]

Recalling my Jr. High pre-algebra classes: distance equals rate times time.

A rate (e. g. 10 kilofurlongs per fortnight) is a speed. Ideally, time is measured in the same unit as the given rate.

If I travel at a rate (or speed) of 10 kilofurlongs per fortnight for 2 fortnights, you multiply 10 by 2, the fortnights cancel out, and you have a total distance of 20 kilofurlongs.

"Rate of speed" is redundant, similarly to "PIN Number", "ATM Machine", and the like.

 

[SageBrush]

A rate (e. g. 10 kilofurlongs per fortnight) is a speed.

The other way around. Speed is a rate.

'x' per 'y'

 

[DopeGhoti]

Yes, "is" is commutative; particularly with maths.

 

[Azathoth]

Just to further roiI the off-topic waters, I would suggest that this event (batteries flying and M3 in pieces) actually DID result from a "high rate of speed" ... rapid deceleration.

 

[Kevin-rf]

Just to further roiI the off-topic waters, I would suggest that this event (batteries flying and M3 in pieces) actually DID result from a "high rate of speed" ... rapid deceleration.

There is also a very good chance he was still accelerating when he impacted the pole.

For some reason I doubt he was like dude, this is the speed to drive at and was traveling at constant speed.

 

[qdeathstar]

the car was traveling at a high rate

a) of comfort
b) of autonomy
c) of efficiency
d) of speed

 

[alexgr]

Just to further roiI the off-topic waters, I would suggest that this event (batteries flying and M3 in pieces) actually DID result from a "high rate of speed" ... rapid deceleration.

So, let me upgrade this conversation to ludicrous.
If we neglect for now (save for plaid conversation) the change of the angular momentum of rotating tires due to the torque exerted by the impact, the maximum velocity of the center of mass of a rotating tire will be 100 mph = 44.7 m/s. If the tire is ejected at 45 degree angle to the ground, we can calculate the height and the distance of where the flying tire reaches its apogee.
Let the horizontal axis be X, and the vertical axis be Y.
X = Vx * t and Y = Vy * t - (1/2)*g*t^2, from the equation for Y, t = (2*Vy)/g is the time before the tire comes back to the ground, so this is the total flight time. Taking this time and the equation for X, X_max = Vx * (2*Vy)/g or considering that at 45 degree angle Vx=Vy =V*sin(45), X_max = 2*V^2 * (sin[45])^2 / g. At apogee, X_apogee = X_max / 2. Now, taking time to the apogee as t/2 and plugging it back into equation for Y, Y_apogee = (1/2) * V^2 * (sin[45])^2 / g.

Plugging in the numbers, X_apogee = 203.8 m = 668.6 ft and Y_apogee = 50.98 m = 167.2 ft. So, considering the Starship height is 50 m, a Model 3 flying wheel could clear its height.

 

[Azathoth]

@alexgr:

Close, but your numbers only apply in a vacuum (no air resistance) ...

 

[pilotSteve]

@alexgr:

Close, but your numbers only apply in a vacuum (no air resistance) ...

.... or in a Monty Python skit

 

[alexgr]

@alexgr:

Close, but your numbers only apply in a vacuum (no air resistance) ...

... so far, I've got that if A=sqrt(m*g/c) and B=sqrt(4*c*g/m), where c is the drag coefficient of the tire (estimated to be 0.045) then
V_y(t) = A*(exp(B*t) - (A-V0_y)/(A+V0_y)) / (exp(B*t) + (A-V0_y)/(A+V0_y))

Now, I have to integrate this to get position Y(t) = integral (V_y(t) dt) and then plug in the time to the apogee
t_apogee = (1/B) * ln([A-V0_y]/[A+V0_y])

I'll take a break... you can check solutions in the meantime.

 

[Kevin-rf]

Why are there dimples on a spinning golf ball? To give it lift. Assuming he had any tread left (a very big assumption) the tire should have generated some additional lift as it sailed into that bathroom wall.

Wonder if we should ask Elon for tires that generate lift like a golf ball...

 

[jjrandorin]

Wow, what a thread!

Everything from a Model 3 torn in half, to flying tires, flaming batteries, Mad Libs, The Doors, meth, frickin’ laser beams and one of the most pedantic arguments I’ve ever seen.

Gotta be an award for this thing!

I have been following this one and it is a doozy. Even though quite a bit of it is off topic, (and the "discussion" happening in this thread is absolutely NOT where I thought the thread would go, nothing rises to the level of needing to put on a mod hat and do anything (other than move other posts about it into this existing thread, anyway).

 

[Kevy Baby]

I have been following this one and it is a doozy. Even though quite a bit of it is off topic, (and the "discussion" happening in this thread is absolutely NOT where I thought the thread would go, nothing rises to the level of needing to put on a mod hat and do anything (other than move other posts about it into this existing thread, anyway).

Your post is technically incorrect as there is only one opening parenthesis but two closing parentheses.

Please do not lend any seriousness to my post - I just wanted to add to the wonderful inanity that is this thread.

 

[qdeathstar]

Your post is technically incorrect as there is only one opening parenthesis but two closing parentheses.

Please do not lend any seriousness to my post - I just wanted to add to the wonderful inanity that is this thread.

why, syntax changes everything.

 

[jjrandorin]

Your post is technically incorrect as there is only one opening parenthesis but two closing parentheses.

Please do not lend any seriousness to my post - I just wanted to add to the wonderful inanity that is this thread.

.... /e must... resist... urge... to edit... my post... and this one too....

(the 1st closing parenthesis was supposed to be after "where I thought the thread would go)

 

[dmurphy]

(the 1st closing parenthesis was supposed to be after "where I thought the thread would go)

SYNTAX ERROR.
LIST
10 PRINT “Missing Quotation Mark.
RUN
SYNTAX ERROR.

 

[jjrandorin]

SYNTAX ERROR.
LIST
10 PRINT “Missing Quotation Mark.
RUN
SYNTAX ERROR.

Lol! Sigh .....

I need to slow down, or put down the whisky I am sipping while browsing the forums.

 

[Azathoth]

Lol! Sigh .....

I need to slow down, or put down the whisky I am sipping while browsing the forums.

Sorry, but it is spelled “whiskey”—with an e—in the United States and Ireland.
No soup (or whiskey) for you!

 

[Kevy Baby]

This thread has been a bright spot in my day.

 

[alexgr]

... so far, I've got that if A=sqrt(m*g/c) and B=sqrt(4*c*g/m), where c is the drag coefficient of the tire (estimated to be 0.045) then
V_y(t) = A*(exp(B*t) - (A-V0_y)/(A+V0_y)) / (exp(B*t) + (A-V0_y)/(A+V0_y))

Now, I have to integrate this to get position Y(t) = integral (V_y(t) dt) and then plug in the time to the apogee
t_apogee = (1/B) * ln([A-V0_y]/[A+V0_y])

I'll take a break... you can check solutions in the meantime.

Well... in process I found that
1) I have missed one negative sign in the Newton's law: m * Y'' = - c * (Y')^2 - m*g
2) I need to include the Magnus force to account for an extra force that can change the trajectory of the tire due to pressure difference on different sides of the rotating tire (a variant of Bernoulli effect).
3) The gyroscopic effect, that is the conservation of the angular momentum of a rotating tire, has to be accounted somehow because the Magnus force may exert torque on the rotating tire.
4) I read several relatively new academic papers on kinematics of tennis balls and frisbees.

... S$!t , this is an active research field! We need more data, I guess ... or not?