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Q: race track "bank" angles ( Answered,   7 Comments )
Question  
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.
Answer  
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
Comments  
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

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