g's on Banked Curves ```Name: Brian S. Status: educator Age: 30s Location: N/A Country: N/A Date: 11/26/2004 ``` Question: Hi, I have a question about the correct way to determine the number of g's a rider in a car traveling on a "banked" curve will experience. Regarding the number of g's felt by a rider on a car, can I just take the centripetal acceleration (ac) and divide it by 9.8 OR........... must I use the normal force? would I set Fn cos(theta) = mg which is..... Fn = mg/cos(theta) then divide by "mg" which would result in, number of g's = 1/cos(theta) which result is correct? Replies: Brian, The first method is correct, provided you know both speed of the car and radius of the track. There is no vertical acceleration: the vertical component of the normal force balances gravity. Provided the car neither speeds up or slows down, all horizontal force is dedicated to making the car turn. This is the centripetal acceleration. If you also have a change of speed, you must combine the forward acceleration with the centripetal acceleration. As they are perpendicular, the Pythagorean Theorem must be used. To use the normal force, you need to determine the horizontal component. As indicated, Fn=mg/cos(theta). The portion of the force that contributes to acceleration is Fn*sin(theta). If there is not much friction, you can say that the net force is F=mg*sin(theta)/cos(theta)=mg*tan(theta). Acceleration is then F/m=g*tan(theta). Still, this assumes that friction has an insignificant effect. Using centripetal acceleration is more accurate. Ken Mellendorf Math, Science, Engineering Illinois Central College Your second method gives you the correct answer. You can see that the first method gives you an incorrect answer by considering either of two limiting cases. Let the speed of the car go to zero or the radius of the curve go to infinity (so the road is straight). In either case the centripetal acceleration is zero, so your first method would say the number of g's felt by the rider is zero. The second method gives the correct answer since theta is zero and so Fn = mg and number of g's = 1, which is surely correct. Best, Dick Plano... Neither. The acceleration of the driver is g downward and v*v/r outward, independent of the bank angle 'th' of the road. However, the orientation of the car under the driver does, of course, depend on the bank angle, so we should recast the accelerations mentioned above into the coordinate system of the car. I will use z for vertical and x for horizontal; and z', x' for the car's coordinate system: ``` ^ ^ ^ -g z = -g (cos(th) z' + sin(th) x') ^ ^ ^ v*v/r x = v*v/r (cos(th) x' - sin(th) z') So the driver feels a "downward" (-z') force of m [g cos(th) + v*v/r sin(th)] and an "outward" (x') force of m [v*v/r cos(th) - g sin(th)] ``` Tim Mooney Both, added together with vector sums. If the car is travelling arbitrarily slowly, your second answer (gravity) is mostly correct. If the car is travelling faster than cars usually can, on a bank nearly 90 degrees, your first answer (centrifugal forces) is mostly correct. At speeds in between, and any time you want an exactly correct answer, you need to do the vector sum of downwards gravity and outwards centrifugal force. ```.............................................. . . . A (centrifugal virtual gravity) . . M---> . . |\ . . G | \ B (vector sum, it's diagonal) . . V \| . . . . G: (earth normal gravity, 9.8m/sec2) . .............................................. ``` M: car, mass of car in kilograms, (doesn't matter, we won't use this number) V: car's speed, in meters/sec R: horizontal radius of road's curve, in meters. G: magnitude of local gravity, G = 9.8 meters/sec2 A: centrifugal virtual gravity, A = V^2/R (your "ac", not yet divided by 9.8) (same as centripetal acceleration, but thought of as opposite in direction.) B: combined apparent gravity B= SQRT[ A^2 + G^2 ] (simple, no cosines because A and G are perpendicular) Th1: angle of B away from vertical G Th1 = arc_tangent[A/G] = arc_cosine[G/B] = arc_sine[A/B] Th2: banking-angle of roadbed, surface angle relative to horizontal If Th2=Th1, the tilt of the banked roadbed is perfectly matched to the speed, and this vector sum B is directly perpendicular to the roadbed, so there is no side-pulling force, and you have your apparent gravity answer. (In general, I like to call this function: A = sqrt[ B^2 + C^2 + D^2 ... ] the "Pythagorean Sum" of B,C,D, etc.) If the road bank isn't matched, you will want to "decompose" the apparent gravity B into "downwards" and "sideways" forces in your tilted frame of reference on the banked roadbed. For this you use sines and cosines. D: downwards part of apparent gravity, normal to roadbed, in meters/sec2 S: sideways part of apparent gravity, parallel to roadbed, in meters/sec2 Th1, Th2: as before ```.............................................. . . . A . . M------> . . |\ | . . G | \ Th3 \ D (drawn poorly) . . | \ \ . . | Th1 \ | Right- ___--- . . v B \ \ angle ___--- banked road . . _\|_--- . . ___--- Th2 . . ___--- ______________________horizontal . .............................................. ``` Th3: relative angle of apparent gravity, relative to road-bed normal. Th3 = Th1-Th2 (or vice-versa) ```.............................................. .Tilted into your perspective . . ....... on the banked road: . . | | . . .\ | D . . . | Th3 | . . . \ | . . . B| | . . . \ | . . . | | . . ___ .----->\V____________banked road bed . . S . .............................................. D = B x cosine(Th3) S = B x sine(Th3) ``` S is of course the force you feel that pulls you sideways in your seat, or makes the car skid-out or roll-over. D is how hard you feel pulled-down into your seat. finally, the part about dividing by 9.8m/sec2: Gd = D/G , answer in "Gee's" (will be larger than 1.00 "Gee" when at speed on a matching banked road) Gs = S/G , answer in "Gee's" (will 0 on a speed-matched bank, + or - otherwise, typically between +1 and -1.) I think that's your complete answer. Hope you can find it practical. We could talk a lot about whether gravity vector formally should point up or down, and the centripetal vector should point inwards or outwards relative to the road's curve. I just used my best common sense, consistently adhering to the perceived direction of apparent gravity-like force. cordially, Jim Swenson I am afraid I must disagree. To say that a driver sitting in a car at rest does not feel any g's is just not correct. You can check by dropping a ball and seeing how rapidly it accelerates downwards. Now consider a car in a roller coaster not going around a corner, but at the bottom of a loop. Assume the centripetal acceleration, which is now in the vertical direction upwards, is g. To say the number of g's the rider feels is 1 g does not make sense to me. If the driver drops a ball, he will see the ball accelerate downwards at 2 g. You can say that one g is due to gravity and the other is due to the centripetal acceleration, but we all know by Einstein's Principle of Equivalence that you cannot tell the difference. So why say there is a difference? The same applies to a car going around a curve. Part of the g force felt by the rider is due to gravity and part is due to the centripetal acceleration (now in a horizontal direction), but again you cannot tell the difference Good question! Best, Dick Plano... You are certainly correct in saying that if the track is not banked appropriately for the speed of the car and the radius of the curve, the total force exerted on the car will not be perpendicular to the track. However, that force will always be in a direction perpendicular to a perfectly banked track (tan(theta) = (v*v)/(gR)) and the magnitude of that force will always be N = mg/cos(theta) and so the number of g's felt by a person in the car will be 1/cos(theta). Whether some of that force is provided by the rails or it is all due to the track is irrelevant. Perfectly banked tracks just help the rails and wheels to last longer. Best, Dick Plano... Click here to return to the Physics Archives

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