Gs felt in banked turn

[ntweisen]

Hello. I’m trying to calculate the gs felt on a roller coaster or race car in a banked turn. This is for the ideal banked curve of angle theta where no friction is required to keep the car from sliding to the outside or inside of the curve.
The equations I’ve been given are:

Radius = (v^2)*tan(theta)/gravity
G’s felt = 1/sin(theta)

I put together a table with 10 degrees, 20, 30 etc. and calculated 5.882 gs for 10 degrees and only 1.07 gs for 70 degrees, which isn’t making sense to me. Can someone please explain to help me understand? Are the equations correct?

http://excelspreadsheetshelp.blogspot.com

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[TheTick]

G is the same regardless of bank. The only difference is the track normal relative to vertical.

F = m * v^2 / r

 

[rb1957]

i think the point about banking the turn is so that the resultant acceleration (the resultant of the certrifugal acceleration and inertia (gravity) is normal to the track.

with theta measured to the horizontal, the vertical component of the normal is N*cos(theta), and the horizontal component N*sin(theta)
the applied accelerations are g in the vertical and v*2/r in the horizontal
so you have v^2/r = N*sin(theta) = (g/cos(theta))*sin(theta) = g*tan(theta)
so r = v^2/(g*tan(theta)) ... not quite your expression ? maybe i'm wrong (wouldn't be the first time, won't be the last), maybe you missed a bracket ?

 

[IRstuff]

Where is the velocity in all of this? That's what generates the sideways g's.


P.s. please avoid skimping on punctuation.

TTFN
faq731-376

 

[rb1957]

"Where is the velocity in all of this? That's what generates the sideways g's." ... that would be the "v" (as in v^2/r)

 

[IRstuff]

Oops, my bad, I didn't see that clearly on my phone display.

Your equation should show the bank angle increasing with increasing velocity, so there's a formulation problem. And, you should look at existing track layouts to correlate your friction factor, like:

http://thirdturn.wikia.com/wiki/Daytona_International_Speedway

TTFN
faq731-376

 

[1gibson]

You may make people nauseous if the make the effect of the turn imperceptible. Don't know if it would be better or worse on a rollercoaster than on a train, it may be more unsettling because they will be expecting to feel the force. On the other hand, people who easily become motion sick probably won't be riding it. Either way, hope this is a relevant tangent. If nothing else the reference to the trains should help kick start your research.

http://en.wikipedia.org/wiki/Tilting_train

Excerpt:"Tilting trains are meant to help reduce the effects of centrifugal force on the human body, but they can still cause nausea, a problem that was widely seen on early "active" tilting trains that exactly balanced the outward force. The effect could be felt under maximum speed and tilt, when the combination of tilting outside view and lack of corresponding sideways force can be disconcerting to passengers, like that of a "thrill ride". Researchers have found that if the tilting motion is reduced to compensate for 80% or less of lateral apparent force passengers feel more secure. Also, motion sickness on tilting trains can be essentially eliminated by adjusting the timing of when the cars tilt as they enter and leave the curves. Systems typically tilt the cars based on a sensor at the front of the train, and the slight delay in reacting to this information leads to a short period of sideways force while the cars react. It was found that when the cars tilt just at the beginning of the curves instead of while they are making the turns, there was no motion sickness. To provide information about the upcoming curves before the front of the train reaches them, a GPS system is used.[4] or faster for new track.[5]"

 

[IRstuff]

Is this for school?

TTFN
faq731-376

 

[chicopee]

Your OP reminds me of physics problems.

 

[zekeman]

"Radius = (v^2)*tan(theta)/gravity"
G’s felt = 1/sin(theta)

Both make no sense as theta goes to zero (no banking), r goes to zero and centripetal acceleration would go to infinity, unless OP means theta is the complement.
AS far as g's felt, as the Tick said it is V^2/rg

I get
Equilibrium eq

(*Mv^/r)*cos@ < mu(Mgcos@+(Mv^2/r)*sin@)+Mgsin@
In the absence of friction, mu=0
(*Mv^/r)*cos@ < +Mgsin@
v^2/(rg)<tan@
The limiting r comes from making the inequality an equality, or
v^2/(rg)=tan@
r=(v^2/g*(cot@)

 

[77JQX]

I think your second equation should be:

G's felt = 1/cos(theta)

 

[Walterke]

G's felt=0 when Tan(90-theta)=g*r/v²

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[rb1957]

other than measuring theta to a different datum, isn't that what i posted back on the 9th (at 10am) ?

"G's felt = 1/cos(theta)" ... this corresponds to a 60deg bank being a 2g turn (which it is)

 

[Ron]

The reason for banking is to reduce the lateral load relative to the normal load....thus reducing the potential to overcome friction and fly off the track. For wheels locked on a track, the reason for banking is to reduce the lateral load on the structure and the car and to keep the passengers in the car! It also reduces the overturning moment that might disengage the wheels from the track.