Gs felt in banked turn
Gs felt in banked turn
(OP)
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?
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 - http://scripting4v5.com





RE: Gs felt in banked turn
F = m * v^2 / r
RE: Gs felt in banked turn
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 ?
RE: Gs felt in banked turn
P.s. please avoid skimping on punctuation.
TTFN
FAQ731-376: Eng-Tips.com Forum Policies
RE: Gs felt in banked turn
RE: Gs felt in banked turn
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_Internatio...
TTFN
FAQ731-376: Eng-Tips.com Forum Policies
RE: Gs felt in banked turn
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]"
RE: Gs felt in banked turn
TTFN
FAQ731-376: Eng-Tips.com Forum Policies
RE: Gs felt in banked turn
RE: Gs felt in banked turn
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@)
RE: Gs felt in banked turn
G's felt = 1/cos(theta)
RE: Gs felt in banked turn
NX 7.5.5.4 with Teamcenter 8 on win7 64
Intel Xeon @3.2GHz
8GB RAM
Nvidia Quadro 2000
RE: Gs felt in banked turn
"G's felt = 1/cos(theta)" ... this corresponds to a 60deg bank being a 2g turn (which it is)
RE: Gs felt in banked turn