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 Subject: race track "bank" angles Category: Science > Physics Asked by: nightrain-ga List Price: \$25.00 Posted: 15 Apr 2005 19:46 PDT Expires: 15 May 2005 19:46 PDT Question ID: 509922
 ```Suppose I had a circular race track five (statute) miles in diameter. Now suppose a vehicle were traveling around this track at 125 kts. How many degrees should the track be inclined (banked) to keep the vertical acceleration forces perpendicular to the horizontal axis of the vehicle? And, how did you figure it out?``` Request for Question Clarification by livioflores-ga on 15 Apr 2005 20:15 PDT ```Could you clarify the unit [kts]? Thank you.```
 Subject: Re: race track "bank" angles Answered By: hedgie-ga on 15 Apr 2005 20:27 PDT
 ```It is a problem involving a vector sum, also known as SEARCH TERM: composition of forces explained here: http://www.tpub.com/content/aerographer/14312/css/14312_35.htm On Fig. 2.2 you see two forces, 5 and 10 lb composed to create a resultant force, inclined at angle alpha, up from the horizontal, where tang (alpha) = 5/10 = .5 so that alpha is about 27 degrees according to trigonometric tables http://www.math2.org/math/trig/tables.htm found under the SEARCH TERM: trigonometric tables In your your case the horizontal force is centripetal force F1= omega * omega * M * radius Other force is gravity, vertical down (opposite of Fig 2.2) F2= M * g so your resultant will be inclined down from horizontal by angle alpha= arctang ( g/ omega square * r) M .. mass of vehicle is canceled out. g is acceleration of gravity (.81 m / s *s ) omega is angular velocity in rad / s * s explained here http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/RotationalKinematics/CentripetalForce.html You of course have to convert your custom units to some consistent standard units, SEARCH TERMS: units SI, miles, knots brings you the conversion tables, e.g. http://www.gordonengland.co.uk/conversion/sidef.htm I cannot do the conversion for you, knots and miles give me headache. Hedgie``` Request for Answer Clarification by nightrain-ga on 16 Apr 2005 08:13 PDT ```Wow, you just overstressed my little pea brain. The velocity of the vehicle on the track (125 kts) refers to "nautical miles per hour," which converts to (for out purposes) 143 "statute miles per hour." But I don't think I got an answer to the question: "How many degrees should the track be inclined (banked) to keep the vertical accelertion forces perpendicular to the horizonal axis of the vehicle?"``` Request for Answer Clarification by nightrain-ga on 16 Apr 2005 20:27 PDT ```I don't think I got an answer. The (revised) question was: "Suppose I had a circular race track five statute miles in diameter. Now suppose a vehicle were traveling around the track at 143 statute miles per hour. How many degrees of bank should be built into the track to keep the vertical acceleration forces perpendicular to the horizontal axis of the vehicle?" The "afterthought" questions were: How many degrees of bank should be built into the track if the diameter of the circular track were three, and four miles?" Assumining the vehicle was still traveling at 143 mph. How you arrrived at these figures is like making sausage, I'd rather not know how you did it.``` Clarification of Answer by hedgie-ga on 16 Apr 2005 20:37 PDT ```Night train You are right - and that demonstrates that your brain functions OK: Often smaller CPU is faster and better: you did not got an answer to everything. So far you got tools and links which can produce the answer. You also got an offer - which you can refuse (by emailing the editors) as follows: You formulate the questions in SI units - and I will give you a very simple formula - which gives you answer for any radius of the track. I may even provide the 3 results (in radians) for three radii (radiuses) of track - of your choice - given in meters (or - to be exact in m/rd -- see note). Note: If you measure angle in radians, and radius im m/rd - then length of arc is in m ( arc= radius * angle ... m = m/rd * rd) as described at length here: ".. radian (rad or rd) a unit of angle measure widely used in mathematics and science. One radian is the angle at the center of a circle that cuts off an arc of length equal to the radius. Since the circumference equals 2 pi times the radius, one radian equals 1/(2 pi) of the circle, or approximately 57.295 779°. Using radians to measure angles seems unnatural at first. However, when angles are stated in radians the constant pi tends to disappear from the equations, and this greatly simplifies calculation. For example, the length of an arc is simply its radius multiplied by its angular measure in radians, and the area of a sector of a circle is simply its angular measure in radians multiplied by half the square of the radius. The radian was defined and named by James Thomson in 1873. Thomson was a mathematics professor.." http://www.unc.edu/~rowlett/units/dictR.html and in general under SEARCH TERM:radius, radian, SI, meter, One more detail: I am leaving today for countryside - where there is no broadband. Will be there two weeks. So, you need to wait 14 days or respond very fast. Sorry about that. Oh: if you take the offer, I will throw in (for free - or possible tip) a link which describes the effect of friction. The answer is than a 'safe band' around the 'best slope' of the track - best means prependicular to the acceleration, right?. Hedgie Hedgie``` Clarification of Answer by hedgie-ga on 16 Apr 2005 20:52 PDT ```I just checked that proper spelling would SEARCH TERM Svejking for word I spelled (over-anglicised as) Swejking Sorry about that too``` Clarification of Answer by hedgie-ga on 17 Apr 2005 00:19 PDT ```Well - I have to go but as a test of the Google technology I entered 9.81 meters per second per second in knots per hour into the search engine and got g in your crazy units: 9.81 ((meters per second) per second) = 68 648.8121 knots per hour which (if correct) would allow me to give you a simple answer without compromising my principles (which include an oath not to convert SI into insane units any more ..) angle of enbankment in degrees = 57 * ( v * v) / ( g * r) valid until about 10 degrees in any CONSISTENT set of units, where V is speed of vehicle r is radius of track (assumed to be circle) so if you enter g given above in knots / hour * hour, and rest in miles and miles/hour - it should work. as a check : as v-->0 angle too goes to 0 -- meaning horizontal surface for bigger angles you just have to use trig tables In my defense, Originally you WANTED TO KNOW how it is derived. May be you can add to a question (for a tip, of course), but can you subtract from the question? ThatbI do not know``` Request for Answer Clarification by nightrain-ga on 17 Apr 2005 09:44 PDT ```Hedgie, many thanx for hanging in there, I admire your patience. Let's re-define the proposition. For another \$25 pls provide just the answers to three questions: Suppose I had three circular racetracks, three, four and five miles in diameter. Now suppose three vehicles were traveling, one on each track, at 143 mph. How many degrees of bank would have to be built into each track to keep the vertical acceleration forces perpendicular to the horizontal axis of each vehicle? Hope you have a great two week countryside vacation. Thanks again for your help. Nightrain (the Liberal Arts guy)``` Clarification of Answer by hedgie-ga on 24 Apr 2005 10:43 PDT ```Hi night train-ga I am back from the countryside. I see that race-car gave you the numerical values. Race car has to know this, of course. Thanks race car. You do not have to worry about sending money - it is all handled automaticaly - electronicaly :-). Seriously, I hope that this nice formula angle = atan ( v * v / g *r) (which actually works for any shape of the track -- you just use r=1/curvature) is worth to you the money you spent and that you are happy with the result. If not -please do an RFC, if so - please do rate the answer. H.``` Clarification of Answer by hedgie-ga on 24 Apr 2005 20:20 PDT ```Re: For another \$25 pls provide just the answers to three questions: Suppose I had three circular racetracks, three, four and five miles in diameter. Now suppose three vehicles were traveling, one on each track, at 143 mph If your speed is fixed to a speed v you simplify the formula thus: you type into Google: 143 mph in meters per second and get 143 mph = 63.92672 meters per second Now formula arctangent( v *v *K /g) becomes arctangent (A * K) where A= v *v /g and K is curvature, so we type into google 63.92672 *63.92672/9.81= and get (63.92672 * 63.92672) / 9.81 = 416.577526 final formula is arctangent(416.57 *K) in degrees= where K is curvature of track in inverse meters at any location so for 5 miles track you enter into Google 1/ (5 miles) in inverse meters and get curvature 1 / (5 miles) = 0.000124274238 inverse meters you use it final formula thus arctangent(416.57 *0.00012427423) in degrees to get arctangent(416.57 * 0.00012427423) = 2.96349488 degrees for 5 mile track curvatures for 3, 4 mile tracks are 1 / (3 miles) = 0.000207123731 inverse meters 1 / (4 miles) = 0.000155342798 inverse meters and so enbankements are arctangent(416.57 * 0.00020712373) = 4.93135468 degrees arctangent(416.57 * 0.00012427423) = 2.96349488 degrees respectively voila enjoy Hedgie```
 Subject: Re: race track "bank" angles From: eppy-ga on 16 Apr 2005 00:18 PDT
 ```The Google calculator converts just about any unit to any other Simply entering "125 knots in miles per hour" in the search bar results in: 125 knots = 143.847431 miles per hour Entering "125 knots in furlongs per fortnight" results in: 125 knots = 386 661.895 furlongs per fortnight```
 Subject: Re: race track "bank" angles From: nightrain-ga on 16 Apr 2005 08:44 PDT
 ```Hey - I know you hate "add on" questions. But, while you're figuring out the bank angle for a circular race track five statute miles in diameter (for a vehicle traveling 143 mph) could you also give me the bank angle under the same conditions if the circular track were three, and four miles in diameter? I owe you big for this one...```
 Subject: Re: race track "bank" angles From: myoarin-ga on 16 Apr 2005 18:13 PDT
 ```Now I know why such tracks are curved; no one has to do any calculations, the drivers just find to right angle "by the seat of their pants". :-)```
 Subject: Re: race track "bank" angles From: hedgie-ga on 16 Apr 2005 20:13 PDT
 ```Surely you are joking Mr. myoarin-ga : "the drivers just find to right angle" by building a race track, crashing, building another - and experimenting ?? In the real life, where there is a friction, there is a range of angles, which will allow the car to make a curve without skidding - and engineers designing tracks and rails do make this calculation, and do specify the 'right angle for a given speed - or vice versa'. eppy, you are right - and the answer "125 knots in furlongs per fortnight" is in my view total Sweiking. What I need is a calculator which converts the question in SI units. Is no commentor going to contribute that? Where is Xargi-ga when you need him?```
 Subject: Re: race track "bank" angles From: myoarin-ga on 17 Apr 2005 12:08 PDT
 ```Sorry, hedgie-ga, as so often, I expressed myself poorly. I meant that if the track is banked up in a curve - like that of a bike race track - then you don't have to do any calculations. If you are afraid that the vehicle can go extremely fast, you can extend the curve up to the vertical and always be on the safe side. I could build my track and be sure it would work before you all can agree on terms ;-) But don't ask me at what angle the car would be, that's where the "seat of the pants" comes in; the driver would find the angle at which he was pressed straight down into his seat. We math-weak types have to fall back on simple solutions, but I am real impressed by all the calculations.```
 Subject: Re: race track "bank" angles From: racecar-ga on 18 Apr 2005 00:06 PDT
 ```It is surprising that hedgie refuses to answer this simple question for the very generous fee of \$25. Here is an answer: horizontal acceleration = v^2/r [v is velocity and r is radius. This is the acceleration due to the fact that the car is traveling in a circle]. vertical accerleration = g [this is the acceleration due to gravity. Actually the car is not accelerating vertically because the downward force of gravity is balanced by the upward force from the track, but... ] tan(theta) = vertical acceleration / horizontal acceleration [theta is the angle by which the track is tilted] So, theta = atan(v^2/rg) That is, hedgie gave a correct formula, though he refused to plug in values. Your values are v = 125 knots = 64.3 m/s r = 5 miles = 8074 m g = 9.81 m/s^2 So, your answer is: theta = 3.00 degrees. You can use the modified formula theta = atan(0.00001676 * v^2/r) to get the values for other speeds and radii, using your units of knots for v, and statute miles for r. The 0.00001676 is in there to account for the acceleration of gravity and because, as hedgie pointed out, the units you are using don't go together very neatly. For radii of 4 and 3 miles, you get 3.75 degrees and 4.99 degrees respectively.```
 Subject: Re: race track "bank" angles From: nightrain-ga on 18 Apr 2005 09:35 PDT
 ```Hey, I finally got my answer(s), thank you. To whom do I send my \$25? Nightrain```