Brain Teasers ? or 50 Years On........... ?

Using Don's figures, there will be 949,910 total negative tests i.e. 1,000,000 - 50,090 positives.

Of these 5 results are "false negatives" so chances of a true negative are 949,905/949,910 i.e. 99.9995%

I get the same result trying to determine the conditional probability p(not A|not B) by using Bayes' theorem again:

p(not A|not B) = p(not B|not A)*p(not A)/p(not B) =

                       = (1-p(B|not A))*(1-p(A))/(1-p(B))

                       = 0.95*(1-1/10000)/(1-0.05009)

                       = 99.9995%

Physics and Applied Maths still feature at A-Level as they did 50 years ago................

A small sledge, mass 3 kg is being pulled up a plane inclined at 20 deg  to the horizontal by a rope parallel to the surface. The sledge is accelerating at 0.6 ms^-2 and the coefficient of friction is 0.4. (Take g = 9.8 ms^-2)

What is the tension in the rope ?

havent fully checked my working so could have made mistake but i get:

T = ma + µmgcos20+mgsin 20

it is left as an exercise to the reader to substitute the actial numbers 

enjoy

ken

For those trying ken's substitution challenge.............

m = 3

a = 0.6

µ = 0.4

g = 9.8

cos 20 = 0.9397

sin 20 = ...........................well, you've got to do some work yourself

Remember, if you are using Excel, it likes its angles in Radians rather than degrees.

A-Level vectors but with an aeronautical application.

Point B lies 142 nautical miles true west of A. The wind is from 240 degrees at a steady speed of 30 kts. An aeroplane passes overhead A towards B at 10:00 hrs, at a true airspeed of 120 kts,

Q1. What true heading should the aeroplane adopt in order to counteract the drift due to wind and thus fly overhead B ?

Q2. At what time should the aeroplane expect to pass overhead B ?

i get a vector at 13.3045° down from the line A-B? there is oviously a better way of expressing this Don ...

you can tell i have yet to take flying lessons from you Don...

enjoy

ken

I haven't checked your geometry yet ken, but the angle looks a bit much to me. Using a couple of approximations I would suggest a wind correction angle/drift of about 7 degrees, ie about half your angle. (I will recheck this later today when I have more time and I will explain my approximations as well)

Headings are expressed in degrees clockwise from North (360) so East would be 090, South 180 etc.

Your heading would be 257 deg (T) - I wouldn't worry about the decimals  (my heading would be about 263)

You probably also picked up the Ground Speed as part of your calculations as well ?

i had tinterpreted the wind direction incorrectly. 7 degrees is indeed correct:

from arcsin  (0.25 * cos 60) = 7.18 degrees

enjoy

ken

I  am pretty sure pilots have a handy gizmo that requires no complicated arithmetic at all. Is it called a plotter or similar? That is how I would do it. Actually, i would use a digital solution.

Form this you will know that I have no idea how to solve this problem unaided...but also that the person who set it probably has easy access to both.

this thread is for intellectual fun... at least thats my view anyhow

enjoy

ken

You can solve the problem by drawing the vectors with a scale ruler and a pair of compasses then measuring the angles with a protractor.

The "gizmo" (as you rightly highlighted) is a Dalton computer, usually made of plastic these days. All it does is allow you to draw the vectors more easily than with the ruler, compasses and protractor.

You can do the arithmetic in your head to an adequate degree of accuracy.

Or you can use the "Sine Rule" and "Cosine Rule" to get "precise" answers.

Or you can buy an "app" - which most aspiring pilots do these day, only to find its not allowed in the exams and you have to learn the principles from scratch !!!! Or better still, a GPS system and auto-pilot and a "user handbook" that's a doddle for a 17 year old, but impossible for a 70 year old to sort out !! - which is why the airlines kick you out at 65.

Like anything, if you do it every day, its easy. If you did it at school 50 years ago but then nothing for 50 years, its a bit of "intellectual fun"  (or not !) - nice one ken !

Well, the question was really about VECTORS, I just used an aeronautical context to give it a bit of authenticity. Any way, I have sketched out below, the basics of the vector drawing that will provide the solution.

Obviously the larger and more professional the drawing, the more accurate the solution will be. But I hope that my diagram and instructions are clear ?

If you click on the image it should take you to flicr and possibly an easier to read diagram

Don,

This is not mentioned in your diagram but presumably point Y is 120 units from A along the line AB.

Rather than getting pencil and protractor, you can use the Sine Rule to solve a triangle of velocities, or other vector diagrams.

The following diagram illustrates this and it is the same as the method used by ken c above

Again, it might be easier to read in flickr

Hi sj, No, i'm afraid not.

Its not really necessary to plot the position of A (the start) or B (the finish). Thedistance AB is 142 nautical miles. The Vector Diagram is drawn to any convenient scale such that the lengths of XY; YZ and ZX represent speeds in Knots (Nautical miles per hour) anddirections in degrees clockwise for North. This is what a vector diagram is all about. Each line represents both a speed (length) and a direction.

ZX is 120 units (True Airspeed)................we don't know the direction until the drawing is complete

XY is 30 units (Wind speed).....................we know the direction because it is given (240 deg from)

The direction of XY is known............we want to go from A to B

Thelength of XY isn't known until you draw the arc centered at Z to cut the line AB.....this action solves the unknowns !!!

XY is considerably less than 120 units and represents the Ground Speed of the aeroplane.

You can then scale this from the vector diagram. It will be around about 95 units long ie representing something like 95 kts

I hope this helps, but if not, just say so and I will try to explain again.

I get 93.08 knots i.e. XY = 120*Sin 22.82/Sin 150 = 240* Sin22.82

This would mean that it would take 1 hour and 31.5 mins to travel 142 nautical miles from A to B so it would pass over point B at 11:31.

Well done sj,

You clearly got the concept. ........Vectors, Sine Rule, Time/Distance/Speed all in one A-Level type question.

Take you back to those halcyon days at school.........................

Here is a photo of one of those "gizzmos" that Lionel referred, showing the solution to the Vector question.

Not my best picture, even in Flickr, but I hope it gives a reasonable picture that you can relate to the Powerpoint drawing I showed earlier.

To do these calculations "in your head" you need a few simple concepts.

• I in 60 Rule
• Maximum Drift (Max Drift)
• Simple rule for resolving Vectors

I'll give you time to think about these and comment, before explaining them (if necessary)

I cannot do anything like this in my head and anyone who can is not to be trusted....run, run for your lives!!!

Better start running Lionel................

1 in 60 Rule

1 unit off track at a distance of 60 units is equivalent to a track error of 1 degree

So

• 5 miles off track at a distance of 60 miles is equivalent to a track error of 5 degrees
• 4 miles off track at a distance of 30 miles is equivalent to a track error of 8 degrees
• 4 miles off track at a distance of 120 miles is equivalent to a track error of 2 degrees
• 30 miles off track at a distance of 120 miles is equivalent to a track error of 15 deg.

You get the idea / It works pretty well up to c. 25 degrees with distances above 10 miles

Remember, this is only an approximation, but at least its a very good one !

Max Drift ..........(maximum drift)

Regardless of the direction of the wind :-

• if you fly approximately at right angles to the wind, you will experience Max Drift (but little or no headwind or tailwind)
• If you fly into the wind, or with the wind, you will experience zero drift (but maximum headwind or tailwind)

An aeroplane flying at 120 kts in a wind of 30 kts will experience a Max Drift of 15 degrees (using the 1 in 60 Rule)

If it is flying at anything other than (approximately) at right angles to the wind, the drift will be less than Max Drift

An aeroplane flying from London to say Newcastle would head approx 360 (North) and at 120 kts would get there in about 2 hours.

If that same aeroplane were to fly in a westerly wind of 30 kts, the aeroplane would drift eastwards with the wind and after 2 hours would be 60 miles east of Newcastle. Not good ! The 'plane simply moves with the airmass (wind) as well as under its own thrust [engine(s)]

Of course, if the aeroplane was pointed 15 degrees west of North, it would wind-up more or less over Newcastle, in a little over 2 hours.