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

Aircraft (1)          ...................................................Aircraft (2)
Mass: 2,000 kilograms (kg)          ........................Mass: 4,000 kilograms (kg)
Total Engine thrust: 4,000 Newtons (N)           Total Engine thrust: 8,000 Newtons (N)
V₁ speed: 65 knots (kt)                        .................V₁ speed: 130 knots (kt)
Take-off run to reach V₁: 750 metres (m)        Take-off run to reach V₁: 1,500 metres (m)
Time taken to reach V₁: 30 seconds (s)            Time taken to reach V₁: 40 seconds (s)

1 nautical mile = 6080 ft and 1 metre = 3.28 ft

At V₁ the aircraft experiences an engine failure and take-off is abandoned.

1. How much work was done to aircraft (1) getting to V₁
2. How much power was used to aircraft (1) getting to V₁
3. How much work was done to aircraft (2) getting to V₁
4. How much power was used to aircraft (2) getting to V₁
5. How much momentum does aircraft (1) possess at V₁
6. How much momentum does aircraft (2) possess at V₁
7. How many times greater is the momentum of aircraft (2)
8. How much kinetic energy does aircraft (1) possess at V₁
9. How much kinetic energy does aircraft (2) possess at V₁
10. How many times greater is the kinetic energy of aircraft (2)
11. Which has the greater effect on kinetic energy, mass or velocity
12. What must be done to the kinetic energy so that the aircraft can be brought to a stop

For work use Joules
For kinetic energy use Joules
For power use Watts
For momentum use kgms^-1

Anybody remember their physics or applied maths ?

Mass is the amount of "stuff" in a body

Weight is a Force = Mass x Acceleration

1 Knot is 1 nautical mile per hour

Joules, Watts, Newtons.........???????...................... you could try "Google"

Don't have nightmares, its only designed to revitalise that old grey matter 

Just try a few at a time, no need to do all 12 at once.

And next time you fly off on holiday or business, don't worry, Boeing and Airbus have sorted all this out, all we have to do as pilots is put the brakes on before V₁ or keep going after V₁

1. 1,120.24 KJ

2. 37.34 KW

3. 8,962 KJ

4. 224 KW

sjbabbey posted:

1. 1,120.24 KJ

2. 37.34 KW

3. 8,962 KJ

4. 224 KW

It's good to see somebody giving it a go SJ, and your answers are in the right "Parish" (rather than the right "Ball-Park"), assuming that my answers are right of course

Bear in mind that pilots like to keep things simple.

Work Done = Force x Distance moved

Power = Rate of doing work = Work done ÷ Time taken

I get nice round(ish) answers for 1 - 4 .................(5 - 12 are less roundish answers)

Although I copied this set of sums from one of my ATPL ground school manuals, it's possible that there are some inconsistencies, so my roundish answers might be too good to be true !

I guess if I round up the figure for V1 for Aircraft 1 from 33.47m/s to 33.5 then the figures will come out much rounder I.e.

1. work done is 1,125kJ

2. Power is 37.5kW

similarly rounding up Aircraft 2 to V1 to 67m/s I then get

3. work done is 9,000kJ

4. Power is 225kW

 BTW I assume that a parish is much bigger than a ballpark.

sjbabbey posted:

I guess if I round up the figure for V1 for Aircraft 1 from 33.47m/s to 33.5 then the figures will come out much rounder I.e.

1. work done is 1,125kJ

2. Power is 37.5kW

similarly rounding up Aircraft 2 to V1 to 67m/s I then get

3. work done is 9,000kJ

4. Power is 225kW

---

 BTW I assume that a parish is much bigger than a ballpark.

That part is spot-on SJ ..................... 

On reflection, I somehow think there is a disconnect (mis-understanding), probably due to my poor English

My solution for Q1 is very simple :-

Work Done = Force x Distance

Force = 4,000 N (given)

Distance at V₁ = 750m (given)

Work Done = 3,000,000 Joules (1 Nm = 1 Joule)

Hmm, I assumed that "thrust" and "force" were not synonymous and therefore needed to calculate the aircrafts' acceleration from a standing start to a velocity of v1 over the distances given. This produced acceleration figures of 0.75m/s/s and 1.50m/s/s which when applied to the masses of the aircraft resulted in forces of 1,500N and 6,000N.

I'm guessing that the disparity between my "net" force figures and those for "thrust" would be accounted for by other forces acting on the aircraft e.g. wind resistance, friction.

Never mind. I think I'll call it a day.

sjbabbey posted:

Hmm, I assumed that "thrust" and "force" were not synonymous and therefore needed to calculate the aircrafts' acceleration from a standing start to a velocity of v1 over the distances given. This produced acceleration figures of 0.75m/s/s and 1.50m/s/s which when applied to the masses of the aircraft resulted in forces of 1,500N and 6,000N.

I'm guessing that the disparity between my "net" force figures and those for "thrust" would be accounted for by other forces acting on the aircraft e.g. wind resistance, friction.

Never mind. I think I'll call it a day.

 

 

That's right, the thrust from the engines has to do a lot of work accelerating the a/c, and as you say, in so doing, part of that work is over-comming air resistance, surface friction, wheel friction etc. The question gave gross figures and I should have made that clear. My apologies.

Your knowledge of the fundamentals of physics and applied maths are clearly still there and a real ability to put them to good practice. Nice work SJ.

Time to tidy up the previous couple of questions.

My tree diagram is shown above for the car crash v people wear glasses question.

"C" represents "Crash" and "NC" represents "No Crash" etc

From the tree diagram there are two branches leading to "Glasses wearers" with probabilities..................(3/10) x (5/9) and (7/10) x (1/3)

giving (1/6) and (7/30)

adding gives (5/30) + (7/30) = 12/30 = 2/5

Alternatively, directly from the question, 5 of the 9  crash victims wore glasses and so did 1/3 of the 21 non-crash victims ie 7.   So there are 5+7 = 12 people who wore glasses.

The probability of picking a glasses wearer at random is 12 in 30 ie 2/5

Finding the probability that a glasses wearer is a crash victim is a bit more difficult to grasp IMHO.  But the required numbers are in my previous post and can be abstracted from the tree diagram.

I'll post them later when I have a bit more time.

Just found a bit of time...................

The probability that a glasses wearer has been a crash victim……

You need to find P(C Ι G) = P(C ∩ G) ÷ P(G)

From before P(C ∩ G) = (3/10) x (5/9) = 1/6
          P(G) = 2/5

So P(C Ι G) = (1/6) ÷ (2/5)
          = (1/6) x (5/2)
          = 5/12

George (*) has decided to build his own speaker !

Naim has stopped production of speakers and George only needs one speaker anyway, since he only ever listens in mono. However, his friend Simon has suggested that the speaker cabinet should have the maximum volume possible (don’t ask me ! I’m not Simon).

George can only afford to buy 18m² of good quality veneered MDF. (he already has the necessary x-over, tweeter, mid and bass and he has enough timber and bits-n-bobs for the internal bracing, stand etc) so he only needs to figure out how to use the 18m² MDF to max out the cabinet volume and as a first approximation is prepared to assume that the MDF has no significant thickness.

The enclosure will be a six-sided cuboid (George will discard the cut-outs for the speaker placements and any ports that he decides to incorporate). For aesthetic reasons George has already decided that the frontal area shall be three times as tall as it is wide.

What is the maximum volume cabinet George can build and what are it’s three dimensions.. Anybody able to help George ?

(*) All names are entirely fictional and any resemblance to real people either dead or alive is entirely co-incidental ........................

Dimensions : 1m * 3m * 1.5m with a volume of 4.5 cubic metres

Don Atkinson posted:

George (*) has decided to build his own speaker !

Naim has stopped production of speakers and George only needs one speaker anyway, since he only ever listens in mono. However, his friend Simon has suggested that the speaker cabinet should have the maximum volume possible (don’t ask me ! I’m not Simon).

George can only afford to buy 18m² of good quality veneered MDF. (he already has the necessary x-over, tweeter, mid and bass and he has enough timber and bits-n-bobs for the internal bracing, stand etc) so he only needs to figure out how to use the 18m² MDF to max out the cabinet volume and as a first approximation is prepared to assume that the MDF has no significant thickness.

The enclosure will be a six-sided cuboid (George will discard the cut-outs for the speaker placements and any ports that he decides to incorporate). For aesthetic reasons George has already decided that the frontal area shall be three times as tall as it is wide.

What is the maximum volume cabinet George can build and what are it’s three dimensions.. Anybody able to help George ?

(*) All names are entirely fictional and any resemblance to real people either dead or alive is entirely co-incidental ........................

This made me smile on so many levels. My second happy moment since finishing work today.

There I was thinking that I had little to contribute or learn from the Forum now I am an official valve admirer for amplification, and my ESLs came home - thus obviating the need to build a speak for myself. Just the one, you see, for the ideal reproduced world of music ... that being mono!

Love you all for making me laugh out loud! See the Carlton has a mate thread.

Also Handel’s Water Music sounded rather well earlier!

Not even Donald cannot Trump that! And I am definitely dead or alive! And not fictitious! 

ATB from George

sophiebear0_0 posted:

Dimensions : 1m * 3m * 1.5m with a volume of 4.5 cubic metres

Spot-on Sophiebear0_0. Well done !

George likes BIG loudspeakers and something 10' x 3'3" x 5' is BIG.....

............even by Focal standards....

Do you want to outline your solution for others ?

In very broad terms I did the following:-

A) created an expression for "d" (depth, front-to-back) in terms of "x" (height)

B) created an expression for "V" (Volume) in terms of "x"

C) differentiated: once to find a max/min value of "x", then twice to confirm the first value was a max

D) substituted my value of "x" in (B) to find the Volume, then in (A) to identify x/3 and "d"

Too many unknowns for this happy old fart! I used to be good at Maths! Now I stick to Arithmetic! summing mostly ...

I am especially good at cycle gearing!

Very best wishes from George

Yes Don - I was very impressed with the size of the cabinet also. But I thought because there was only one speaker to accommodate, it wouldn't look too conspicuous !

I used pretty much the same method as you outlined. I seem to remember that we had a similar problem at the beginning of this thread regarding minimising boundary fence for a car park ? As you mentioned, you need to look at the second differential to determine the sign and then work out whether it is a maximum or minimum.

Regards,

Peter

I have outlined my calculation above. Dv/dX if you recall is the terminology we used for differentiation. I used "Y" for the depth of the cabinet to avoid using "d" so as to avoid confusion with the terminology used in differentiation.

The rest should be straight forward....................I think

George (*) has decided to prepare some custard pies to throw at David Cameron, Jeremy Corbyn and Nicola Sturgeon at their first joint “Let’s Stay” rally next week which kicks off in Warsaw.

He has decided that the best British Custard Pie has a volume of 1000cm³ (sorry George, but I think we might still be stuck with this half-hearted European metric system whether we are In or Out !). The baking tin will be circular, diameter “d” and “t” tall.

George naturally wishes to economise on the Welsh tin-plate he needs to use in the fabrication of the baking tin. Given that the cost is the same per sqcm regardless of the shape, he decides to buy two pieces, one circular (for the base) and the other, rectangular (for the circumference).

Ignoring overlaps for the joints, how much tin-plate will he need to buy as a minimum and what dimensions should he order for each of the two pieces ?

(*) All names are entirely fictional and any resemblance to real people either dead or alive is entirely co-incidental ........................

Baking tin:

Rough calculations suggest 440 sqcm of tin. Height "t" 6.83cm. Radius also 6.83 cm, so diameter "d" 13.66cm

Don! You really are a naughty boy! But your posts lighten my day. I thought the Tin Mines were in Cornwall!

Today I serviced the Bottom bracket on a fifty year old Claude Butler cycle of exquisite condition, though the set-up needs finessing ...

I added to the mix a beautiful gipiemme crankset of the era just for the pleasure of matching one classic with another!

I am sure people are bored by my appreciation of British made cycles from the pre-EEC era! 

I can be naughty also!

ATB from George

And my gipiemme crankset is also rather suitable on this British Classic bike. Albeit that the part was made in Italy, decades ago!

ATB from George

My Carlton looks rather austere beside the flamboyant Butler. But it really is good cycle, Rather like Bentley claimed for their Speed Six [6 plus litre] sports car. Good from six milea an hout to ninety-six on top gear!

I have exceeded forty on the Carlton with absolutely stability!

ATB from George