Furry Dice and Vectors

Bishop

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Hi guys

I've been looking at vectors - the theory of

One thing I'd like to ask... If I have furry dice hanging from my rear view mirror, and notice they are now hanging at a 45 degree angle while turning around a corner, does that mean the down vector (gravity) is exactly equal to the side vector (centifru-whatsit) so I'm making a 1 G turn?!

Cos that just sounds too bizarre to be true :nono:

- B
 
It checks out: yes, you're pulling 1G if you're on level ground. But...just what kind of turn are you pulling? :blink:
 
Hi guys

I've been looking at vectors - the theory of

One thing I'd like to ask... If I have furry dice hanging from my rear view mirror, and notice they are now hanging at a 45 degree angle while turning around a corner, does that mean the down vector (gravity) is exactly equal to the side vector (centifru-whatsit) so I'm making a 1 G turn?!

Cos that just sounds too bizarre to be true :nono:

- B
Depending on the car's suspension, it's also possible that the body is "rolling" somewhat to the outside of the turn, which can make the fuzzy dice hang at a 45 degree angle to the mirror when they are at a less than 45 degree angle to the ground.
 
one of the reasons i put the dice in the xr2 is because i wanted a visual g-meter like this, to illustrate the forces acting on the craft. Of course in practice now they are static unfortunately.

The effects of your car cornering would be the same as if you were translating sideways and up simultaneously in the xr2, and the dice would respond in the same way by pointing down and to the side at 45 degrees.
 
Depending on the car's suspension, it's also possible that the body is "rolling" somewhat to the outside of the turn, which can make the fuzzy dice hang at a 45 degree angle to the mirror when they are at a less than 45 degree angle to the ground.

Yes, but I doubt the car would "roll" 45 degrees...

Then again I know virtually nothing about car suspension.

one of the reasons i put the dice in the xr2 is because i wanted a visual g-meter like this, to illustrate the forces acting on the craft. Of course in practice now they are static unfortunately.

Don't they have something akin to dice on Soyuz for that?

Also, is there not any possibility of the dice wondering around on their tether and whacking someone in the face, or whacking someone in the face during acceleration? It's not bound to do any damage, of course- but I'd imagine it could become rather startling/annoying.

Either way, it's a pity the dice are static.
 
Are those dice hanging there still or are they wobbling about?

It might have been just inertia swinging them up there - they aren't massless after all. You'd need to then average their angles to get a usable figure.

But then again, 1 g isn't that much - that's doing a 90° turn while going 20 km/h in sqrt(2)*20/9.81 s ≈ 3 seconds.
 
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Yes, but I doubt the car would "roll" 45 degrees...

Then again I know virtually nothing about car suspension.
No, but if the car has rolled only 15 degrees (when you have a floaty suspension like mine, that's easy), you'd only need to be making a ~.6 degree turn in order to get the dice to hang out at 45 degrees relative to the mirror.

But then again, 1 g isn't that much - that's doing a 90° turn while going 20 km/h in sqrt(2)*20/9.81 s ≈ 3 seconds.
Assuming that the car is fully controlled at the time, maintaining a 1g turn for any length of time is actually fairly impressive.
 
1g in a road car is impressive, true. However my head is spoilt to such things through years of watching F1:

A newer and perhaps even more extreme example is the Turn 8 at the Istanbul Park circuit, a 190° relatively tight 4-apex corner, in which the cars maintain speeds between 265 and 285 km/h (165 and 177 mph) (in 2006) and experience between 4.5g and 5.5g for 7 seconds—the longest sustained hard cornering in Formula 1.

But the F1 car is in a different league again: 6g has been recorded at Suzuka’s 130R corner and 4-5g is normal in fast turns.
 
Yeah, I just realized that a 1g turn on a flat road is an equivalent of the car standing still on a 45° slope... is that correct?
 
Yeah, I just realized that a 1g turn on a flat road is an equivalent of the car standing still on a 45° slope... is that correct?
Not quite. Same direction (from driver's perspective), but different magnitude. It's an odd thought (especially when thinking of higher G turns) but the 1G turn is a good deal more force than with the slope. More force holding you to the road, but also more force pulling to the side. :P Seems dangerous indeed. ;)
 
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Sums work

But then again, 1 g isn't that much - that's doing a 90° turn while going 20 km/h in sqrt(2)*20/9.81 s ≈ 3 seconds.

That's more or less how it happened Tori, making a 90deg turn at low speed :thumbup:

May I ask where you found that formula? I could put it in Excel and save some tryre marks on an empty car park :lol:

- B

---------- Post added at 09:48 AM ---------- Previous post was at 09:45 AM ----------

Seems dangerous indeed. ;)

Too right Nick, 1G turn seemed a lot, which is why I checked it with you guys

Thanks all who replied! :tiphat:

- B
 
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You take the velocity vector before and after the turn and compute the difference. In the case of a 90° turn the difference is a diagonal of a square, i.e. delta_v = sqrt(2)*speed.

Edit: or if your speed before and after the turn is the same, you can use delta_v = 2*sin(angle/2)*speed

Now if you want 1 g of sideways acceleration you divide the delta-v by that (9.81 m/s/s) and what comes out is the time you need to do it in.
 
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Except to get the right units, you need to insert speed in m/s, not km/h. You calculated acceleration when going 20 m/s, and that is 72 km/h :p

Also, I have another formula.
Let's assume your path is a fragment of a circle when turning by an angle of [math]\alpha[/math] in [math]t[/math] seconds with velocity [math]v[/math]. You will then travel arc of length [math]vt[/math], which happens to be equal to [math]r\alpha[/math] ([math]r[/math] being the radius of the circle).

Hence, [math]r = \frac{vt}{\alpha}[/math].

Centrifugal acceleration: [math]a = \frac{v^2}{r} = \frac{v^2\alpha}{vt} = \frac{v\alpha}{t}[/math]

So, when turning 90 degrees ([math]\frac{\pi}{2}[/math] radians) with velocity 20 km/h = 5.5 m/s for 3 seconds, your acceleration equals a = 5.5*(pi/2)/3 = 2.88 m/s^2 = about 0.3 G.

With velocity 20 m/s, you get 10*pi/3 = 10.47 m/s^2 = about 1.07 G.
 
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i can't help but to wonder...

how does one "hang" anything without effective gravity? :rolleyes:

i mean... if no force pulls "down" on the dice, it isn't really hanging as much as just floating around :lol:


sorry... :threadjacked: carry on :thumbup:
 
i can't help but to wonder...

how does one "hang" anything without effective gravity? :rolleyes:

i mean... if no force pulls "down" on the dice, it isn't really hanging as much as just floating around :lol:


sorry... :threadjacked: carry on :thumbup:
Force of gravity is countered by a centripetal force. (I hope that's an okay way of using that word :lol:)
 
Fizyk, oh man, you're right. That is so embarassing. Especially since I'm the guy who always goes "neener neener wrong units, dimensions don't check out". Darn it.

:embarrassed:
 
i can't help but to wonder...

how does one "hang" anything without effective gravity? :rolleyes:

i mean... if no force pulls "down" on the dice, it isn't really hanging as much as just floating around :lol:


sorry... :threadjacked: carry on :thumbup:
You don't "hang" it, you "mount" it.
 
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