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Suggest merging this article, cant (road/rail) and cant deficiency into one article "Banking / Cant (engineering)"
FengRail ( talk) 21:27, 11 April 2009 (UTC)
The figure banked_turn.png is incorrect and should be removed or corrected. First, the figure shows both a centripetal and a centrifugal force acting on the same body, which cannot happen in any frame of reference. Second, the "vector sum of body forces" is shown as the resultant of the weight and the centrifugal force, and shown to be equal and opposite to the lift (and generally downward, which is obviously false). The only real forces acting on a plane are the weight and the lift, the latter larger by cos theta, and their resultant is the the centripetal force, directed horizontally. This figure could be corrected if the pink and red arrows were removed, and the yellow arrow shown as the sum of the green and blue.-- 74.104.101.121 ( talk) 03:08, 23 December 2010 (UTC)
The idea of the diagram was originally to illustrate the nomenclature of the various forces that may come to play in such scenario, but I agree that the result can be confusing to the uninitiated.
The vector sum of all body forces is a leftover of the
diagram from which the one in question has been derived (the original purpose was to show why the lift has to increase) and it is correct, as the only
body forces (either real or apparent) are weight and centrifugal force, but that's irrelevant anyway.
I'm happy to edit the picture and remove centrifugal and purple resultant. I think the yellow resultant should be left though, as it is explicitly mentioned in the article (and indeed was agreed to be added to the first draft of the diagram).
--
Giuliopp (
talk)
17:52, 23 December 2010 (UTC)
I see the picture has been edited. I would make one more request for editing the picture. Right now it shows three vectors, green for lift, purple for weight, and yellow for centripetal acceleration. There are a number of statements suggesting that there are only two forces, lift and weight, on the aircraft. I agree. The lift can be broken down into component vectors perpendicular and parallel to the lift vector. The centripetal acceleration is just the component of the lift vector perpendicular to the weight, meaning it is in the horizontal (constant altitude) plane. The other component is the one parallel to the weight, and it is not shown. I suggest two changes - first, show the components as dashed lines to indicate that they are just components of a force (lift), and second, indicate clearly that the component of lift opposite the weight must balance the weight out completely in order to maintain level flight. This will show the "centripetal force" as essentially an unbalanced force that is turning the aircraft in its circle. — Preceding unsigned comment added by SaintLouisEE ( talk • contribs) 23:48, 10 September 2022 (UTC)
The first two equations in this section are wrong. Note that the equation "N cos(theta) = mg" should be "N = mg cos(theta). The posted equation implies that the normal force is always equal to or greater than the weight. If this equation holds true then as the bank angle approaches 90 degrees, the normal force approaches infinity, which is absurd.
The equation "N sin(theta) = v^2/r" is also wrong. It should be "mg sin(theta) = m v^2 cos(theta)/r".
The remaining equations are correct, however, the text needs to be changed to reflect the contents of the correct equations. Chasboson ( talk) 19:08, 26 June 2014 (UTC)
The normal force does not increase with bank angle. It decreases. The equation for normal force has nothing to do with velocity. Neither "N = cos(theta)" nor "N cos(thata) = mg" is a function of velocity. It is the same on a vehicle resting on the road as on one moving. What does change is the centrifugal force. The task here is to balance that force with the component of the weight (not normal force) opposing it. I agree that at 90 degrees the object will fall, but that's because the normal force drops to zero and the weight prevails.
Chasboson ( talk) 21:28, 26 June 2014 (UTC)
As I stated, the normal force has nothing to do with motion along the curve. The centrifugal force does. Evidently you either haven't read the page on "normal force" on Wikipedia, or you haven't had time to 'correct' it to match your formula. You certainly won't find the equation "N cos(theta) = mg" or the equivalent "N = mg/cos(theta)" there. I see a real mismatch here and I think it would be important to reconcile the differences. I've derived the equations for motion along a banked curve using the conventional and accepted formula for the normal force and arrived at the correct answer. The version here has two errors which cancel and still get the correct answer, but I don't think that is acceptable.
Chasboson ( talk) 00:42, 27 June 2014 (UTC)
It's understandable how these cases can be confused as they look similar, but if you look closely the difference in both the maths and geometry is clear. To save me having to draw pictures here are two diagrams.
In both diagrams the forces form a right angled triangle. But the triangles are different.
Again this corresponds to the physics. On an inclined plane as the angle increases the normal force decreases as more and more of the weight is supported by friction. But on a banked curve as the slope increases the normal force goes up as the centripetal force increases. On e.g. a racetrack this corresponds to the steepest banked curves being at the sharpest/tightest corners. The sharper the corner the more centripetal force is required to make the turn, requiring more banking and more normal force.-- JohnBlackburne words deeds 14:53, 27 June 2014 (UTC)
Well, it appears that our disagreement is over the definition of the normal force. I maintain that the normal force is found from "N = mg cos(theta)". You claim that the normal force is the sum of the gravitational component and the centripetal component. Where you write "N^2 = mg^2 + (mv^2/r)^2" I would write "(mg)^2 = N^2 + (mg sin(theta))^2". So if 'N' is affected by lateral forces, this may be a good place to introduce citations. In my physics texts the normal force is found by resolving the weight into two components: one along the surface and one perpendicular to it --- the perpendicular component is 'N', or rather the reactive perpendicular force is 'N'. Other forces may affect the vehicle, but they are in addition to and separate from the normal force. At least that is what I find in my texts.
Chasboson ( talk) 18:28, 27 June 2014 (UTC)
There remains the difference between the two equations "N cos(theta) = mg" and "N = mg cos(theta). They can't both be right. I understand that you 'changed" the triangle, but didn't explain why. It certainly isn't necessary, because the desired result follow directly from the equation "N = mg cos(theta)". Moreover, changing the triangle causes the balancing of forces to occur along the horizontal axis rather than along the incline. Using "N = mg cos(theta)" results in the balancing to occur along the banked surface, which makes more sense. Under this approach the gravitational force pulling the vehicle downward along the surface is "mg sin(theta)" and the centripetal force pushing it up along the curve is "m v^2 cos(theta)/r". Equating these and dividing both sides by "mg cos(theta)" results in the desired relation. Simple, neat and unambiguous.
Interestingly, the normal force 'N' doesn't even appear in the result.
Chasboson ( talk) 15:56, 28 June 2014 (UTC)
Hmmm. I realize the problem involves a frictionless surface. That's why the lateral forces need to cancel. I thought your first drawing above illustrated the forces nicely except that without friction the component of centrifugal force balances the component of gravitational force.
I'm willing to leave it at this point. I think there is a better way to derive the solution to the frictionless banked curve problem, but suppose getting the right answer is the main thing.
Chasboson ( talk) 18:43, 28 June 2014 (UTC)
It seems to me that taking the equation for 'F' and setting F to zero is the simplest solution to the original frictionless banked turn problem. The desired result can be found immediately with a little algebra from that. Doing it this way eliminates any need for the normal force. That makes sense because the normal force does not generate lateral components.
Chasboson ( talk) 17:59, 29 June 2014 (UTC)
I decided to make one last post to recap and clarify my objections to the derivation of the 'frictionless banked turn' equation. On rereading this thread it is clear that misunderstandings, skipping around and switching topics has interfered with any cogent discussion. I want to focus on very specific points.
First, the equation:
My complaint was that the normal force, as defined by this equation, will always equal or exceed the weight. As the bank angle approaches 90 degrees, N approaches infinity. I claimed that this cannot be correct. There is no mechanism for an object, solely under the influence of gravity, to exert forces of 2 time its weight or 5 times its weight etc. That's more like sorcery than physics.
When I first posted my objection, the response I got had to do with a 'wall of death' and various references to motion and velocity. However, this equation contains no term with a velocity component. It is unaffected by velocity because once the weight and bank angle are given, the normal force, N, involved is determined. It is fixed by those values and if the equation was correct, N would be the same for zero velocity or at the speed of light. I noticed that different equations for N were later invoked which included centrifugal force and which can increase N beyond the gravitational component, but no other forces are present in the equation under discussion, which merely purports to show the relation among three variables, none of which is velocity. Incidentally, for this equation it doesn't matter if the surface is frictionless or not. If sufficient friction is present, the object will exert the force and remain motionless. If the surface is frictionless, it will exert the same force and accelerate down the incline. It makes no difference if the object is in motion or not. The correct equation for the gravitational component of N is
I also complained about the equation
This shows that the forces are being treated along the horizontal axis instead of along the surface of the banked turn. The logical place to resolve the forces is along the incline which is where the object is constrained to move. This is shown in the first figure above. The response I got to the objection was that it is a frictionless surface and "there are no forces" along the banked surface. I claim that there are two forces whose sum is to be set to zero. Note that their sum is zero, not that the forces disappear. If either of these forces disappears, the other will determine the acceleration of the vehicle up or down the incline. The gravitational force along the incline is
The problem to solve is simply to find an equal and opposite force countering the gravitational force as is done by friction in the first figure above. Such a force, on a frictionless surface, would be available from the component of centrifugal force which lies along the incline, namely:
This is exactly what was done in the recently posted equation for F. Equating these two forces and dividing both sides of the equation by
gives the desired result:
No detour through normals or switching triangles is necessary.
Other than that, the rest of the section is OK.
Chasboson ( talk) 16:41, 8 July 2014 (UTC)
I don't doubt that there is misunderstanding on my side. But I don't think you ever explained how an object, solely under the influence of gravity, can generate forces greater than its weight. I gather you expect the centrifugal force to explain this, but it is not present in the equation I disputed. Neither is any term involving velocity. If you had been able to explain it, this thread would have stopped a long time ago. On the other hand, I understand that it is not your job to educate me, so I accept your frustration in responding. Still, I would like to see an explanation, as this would certainly improve the article.
Chasboson ( talk) 15:57, 9 July 2014 (UTC)
Did you accept the equation for F above as correct? It appeared to me that you did. I certainly do, and this may be the key to agreement, and the conclusion of this thread.
Chasboson ( talk) 20:05, 9 July 2014 (UTC)
OK. I am no longer confused about your exposition. I understand why you dealt with the problem the way you did. However, I completely disagree with you. Thus, there is no longer any need to continue this session.
Thank you for maintaining civil discourse throughout. It is not only a Wikipedia policy, but brings honor to scientific debate.
Chasboson ( talk) 16:40, 10 July 2014 (UTC)
Sorry I wasn't specific. I believe that the appropriate place to zero the forces is along the incline, where the vehicle is constrained to move. You said that was not appropriate and prefer to zero the forces elsewhere. That's a simple, specific disagreement which doesn't allow for compromise. Hence, my statement about terminating the session. I do agree that a three against one vote certainly trumps the one.
Chasboson ( talk) 16:02, 11 July 2014 (UTC)
In an inclined plane : the normal force N =mg cosQ where Q is angle of inclination. But as the law of banking of road says vertical component of normal force N cosQ=mg.The substituting the value of N we get mg cos^2 Q=mg! How can it be possible? Shivaram Raut ( talk) 07:12, 20 February 2017 (UTC)
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I stumbled across this thread and believe the debate (argument) over the use of the normal force symbol 'N' can be resolved. In earlier physics, involving static forces, it is usual to state that N = mg cos(θ). This can elicit a classic difference between mathematicians and physicists because a mathematician is likely to regard this as a definition of N, rather than a current value. In such a case, it should be redefined at the beginning of this article. The confusion would not have arisen if 'N' was identified as a TOTAL force, which may be obvious to some, but which could be clarified by writing: N = Nt = Ng + Nv (vector addition). This way, Ng would be identified as the component of the normal force due to gravity and Nv would be identified as the component due to velocity.
I hesitate to suggest that anyone is 'right' or 'wrong' here, I just think some clarification would resolve the positions taken and would improve the article. — Preceding unsigned comment added by Reslox ( talk • contribs) 18:15, 3 July 2019 (UTC)
I liked the development of the banked turn. I think the approach of starting out with a basic frictionless banked turn is a good way to teach this topic. I would like to suggest that we add a bit more on the aeronautics to describe the forces and what is necessary to achieve a banked turn. What is said in the text explaining that an aircraft needs to use its elevators or put its nose up to create more lift is correct, although we should add that any method that increases lift will do. The major piece that is missing here on the aeronautics side, although it goes a little beyond the banked turn, is that generally when lift is increased, drag is also increased. For the aircraft to keep at a constant speed an altitude, it is necessary that the aircraft apply more thrust to counteract the extra drag. The net result of this is that the engines need to increase their power to achieve the constant speed turn. I am hoping that someone who is an aerodynamicist will add in some description of this. One of the neat things about doing this is that, when in another article, someone wants to understand the flight envelope of an aircraft, they will be able to visualize the limits of maneuver capability - if you don't have enough excess thrust available in your engines to counteract the excess drag from the higher lift mechanisms needed for a banked turn, you will not be able to succeed in the constant speed maneuver. This would also be enhanced by a force diagram that would include drag as well as the lift and gravity vectors already in the figure. SaintLouisEE ( talk) 23:27, 10 September 2022 (UTC)