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
nightrainga
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
nightrainga
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
hedgiega
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
hedgiega
on
16 Apr 2005 20:52 PDT
I just checked that proper spelling would
SEARCH TERM Svejking
for word I spelled (overanglicised as) Swejking
Sorry about that too

Clarification of Answer by
hedgiega
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
nightrainga
on
17 Apr 2005 09:44 PDT
Hedgie, many thanx for hanging in there, I admire your patience.
Let's redefine 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
hedgiega
on
24 Apr 2005 10:43 PDT
Hi night trainga
I am back from the countryside.
I see that racecar 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
hedgiega
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
