Why enlarge Heathrow Airport when you can make it smaller?

paulears

Free Member
Jan 7, 2015
5,653
1,661
Suffolk - UK
http://www.ukbusinessforums.co.uk/forums/showthread.php?t=192697

I thought I remembered his name - his last tinpot project when on and on, like this one - with dodgy physics, mangled facts, misunderstood theories. His Nerissa Glider concept was proven time and time again to be fatally flawed, and now he's totally misunderstood how aircraft operate. Basic aerodynamic functions are forgotten. Missed landings are not uncommon and then a landing becomes a take-off, and if you've used up the short landing runway, where do you go?
 
  • Like
Reactions: Subbynet
Upvote 0

paulears

Free Member
Jan 7, 2015
5,653
1,661
Suffolk - UK
I wonder how the passengers would feel subjected to the rapid deceleration such a system would require? Ever seen what happens to the passengers in a larger plane landing on a carrier? It's dramatic, unpleasant and if you do it to civilians the medical claims will roll in. Whiplash claims by the bucketload. Even if such a crazy scheme was possible (which it isn't) then the development and testing phase would be years - and there would be no financial advantage to airports or the plane manufacturers. The forces put up with it because you can't make carriers bigger. The cost of this system, would be very unattractive to investors, the legislators and the passengers.
 
Upvote 0

Jeff FV

Free Member
Jan 10, 2009
3,891
1,861
Somerset
Swisaw:

I can't dispute your enthusiasm for physics and engineering (I, too, remember one or two of your other threads) but I do think that your understanding of physics is not as full as it might be. This is an observation, not a criticism, - I feel privileged to have had an excellent schooling, followed by a University degree in Engineering. I appreciate that I am very lucky to have had these opportunities.

Can I suggest that you undertake the Introduction to Physics course on Udacity - these are free, online courses at around 1st year undergraduate level (for the beginner courses) - I genuinely think that you will find it interesting (caveat: I haven't done this specific course myself, but have done other Udacity courses, largely for my own interest)

I do not mean to be patronising - I offer the above advice in the spirit of helpfulness. Like me, you seem to enjoy finding out why stuff works and I do think that you would enjoy the course.

Best wishes

Jeff
 
  • Like
Reactions: roydmoorian
Upvote 0

Swisaw

Free Member
Sep 24, 2010
1,849
149
London
Swisaw:

I can't dispute your enthusiasm for physics and engineering (I, too, remember one or two of your other threads) but I do think that your understanding of physics is not as full as it might be. This is an observation, not a criticism, - I feel privileged to have had an excellent schooling, followed by a University degree in Engineering. I appreciate that I am very lucky to have had these opportunities.

Can I suggest that you undertake the Introduction to Physics course on Udacity - these are free, online courses at around 1st year undergraduate level (for the beginner courses) - I genuinely think that you will find it interesting (caveat: I haven't done this specific course myself, but have done other Udacity courses, largely for my own interest)

I do not mean to be patronising - I offer the above advice in the spirit of helpfulness. Like me, you seem to enjoy finding out why stuff works and I do think that you would enjoy the course.

Best wishes

Jeff

Thanks Jeff. Can you do me a favour by checking what is wrong with my figures as a response to Subbynet. If my last figure, 505.238billion joules, is correct could you find its monetary value based on my last gas bill, which was £218.66 for 525 gas units converted to 5,831.11 KWh.

Thank you


Quote:
Originally Posted by Subbynet
You're calculations are wrong, your hydraulics might transmit say 75%-90% efficiency but your mechanical parts will loss some of that energy.

Lets calculate it with no loss and use rough numbers for ease of sums.

1 Megajoule is equal to 1 ton moving at 100mph. So a 500 Tonne jumbo jet using your system generates around 500 Megajoules.

If your rope is 1 tonne in weight, it takes roughly roughly 6-15MJ to create steal (per kilo), so lets say 8MJ per kilo for steel, 1 Tonne = 1000 kilos = 8 Gigajoules. 1 landings every minute as we calculated earlier in the thread means 12 replacements per day at Heathrow, so that equals 96 Gigajoules energy used to create parts.

So you've created 50 Gigajoules using your recovery system, but used 96 Gigajoules.

That's rough numbers, and in reality there would be much more energy used to create the parts.

My Response:


Let us forget about speculation about changing ropes.

Let me to give you real net energy gained from landing one of your 500tonne jumbos:


500tonne= 500k kg.
100mph=160934.4m/h=45m/second


Energy= ½ x mass x squre of speed= ½ x 500,000 x (45 x45) = 506.25Billion Joules, generated by the brake of the jumbo plane.


Energy taken by the rope on both directions= ½ x 1000 x (45/45) x 2=2.03 Mega Joules. The rope has to go back after the plane stops.


So roughly each landing generates 506.25billion - 2.03= 505.238billion joules

__________________
 
Last edited:
Upvote 0
Swisaw:

I can't dispute your enthusiasm for physics and engineering (I, too, remember one or two of your other threads) but I do think that your understanding of physics is not as full as it might be. This is an observation, not a criticism, - I feel privileged to have had an excellent schooling, followed by a University degree in Engineering. I appreciate that I am very lucky to have had these opportunities.

Can I suggest that you undertake the Introduction to Physics course on Udacity - these are free, online courses at around 1st year undergraduate level (for the beginner courses) - I genuinely think that you will find it interesting (caveat: I haven't done this specific course myself, but have done other Udacity courses, largely for my own interest)

I do not mean to be patronising - I offer the above advice in the spirit of helpfulness. Like me, you seem to enjoy finding out why stuff works and I do think that you would enjoy the course.

Best wishes

Jeff

You should have been a diplomat, maybe you are.
 
  • Like
Reactions: Jeff FV
Upvote 0

Jeff FV

Free Member
Jan 10, 2009
3,891
1,861
Somerset
Thanks Jeff. Can you do me a favour by checking what is wrong with my figures as a response to Subbynet. If my last figure, 505.238billion joules, is correct could you find its monetary value based on my last gas bill, which was £218.66 for 525 gas units converted to 5,831.11 KWh.

Thank you

I think you mean 505,000,000 Joules for an aircraft landing

(based on your figures: KE = 1/2 m x v squared:

1/2 x 500,000 x 45 x45 = 506,000,0000

505,000,000 Joules = 140.3 KWh (I used Wolfram Alpha for this calc)

So if you spent £218.66 for 5,831.11 KWh then 140.3 KWh is worth about £5.26

I haven't read the whole thread in detail, but I get the impression that you are trying to harness the energy 'given up' in slowing a jet to a stop on landing. If the figures above are correct, given a near 100% efficiency (nigh on impossible) you are only going to generate the equivalent of £5 worth of gas for each jet that lands.

I must confess, the figures seem lower than I might have thought, there might be an error in there somewhere.

J
 
  • Like
Reactions: Swisaw
Upvote 0

paulears

Free Member
Jan 7, 2015
5,653
1,661
Suffolk - UK
Have you also seen how many people it takes to change the cables, and how dangerous they can be? Aircraft are actually very fragile. Remember the problems with certain aircraft a few years ago when some just appeared to fall apart in the sky, and they ended discovering micro fractures from the edges of the window frame caused the roof to peel off in flight. Carrier landing aircraft have airframes with much more severe time limits on them. Some 747s flying today have amazing hours clocked up, carrier aircraft last a fraction of the time.
 
Upvote 0

Swisaw

Free Member
Sep 24, 2010
1,849
149
London
I think you mean 505,000,000 Joules for an aircraft landing

(based on your figures: KE = 1/2 m x v squared:

1/2 x 500,000 x 45 x45 = 506,000,0000

505,000,000 Joules = 140.3 KWh (I used Wolfram Alpha for this calc)

So if you spent £218.66 for 5,831.11 KWh then 140.3 KWh is worth about £5.26

I haven't read the whole thread in detail, but I get the impression that you are trying to harness the energy 'given up' in slowing a jet to a stop on landing. If the figures above are correct, given a near 100% efficiency (nigh on impossible) you are only going to generate the equivalent of £5 worth of gas for each jet that lands.

I must confess, the figures seem lower than I might have thought, there might be an error in there somewhere.

J

Many thanks Jeff.

Actually I meant 505,000,000,000 Joules by mistake. This seemed displeased Subbynet. This happened because I was using computer calculator, which was giving 20125 for 45 x 45 but who believes that? This morning I did the calculation by hand and discovered the mistake. Strangely now the computer calculator doing it right.
 
Upvote 0

Swisaw

Free Member
Sep 24, 2010
1,849
149
London
Have you also seen how many people it takes to change the cables, and how dangerous they can be? Aircraft are actually very fragile. Remember the problems with certain aircraft a few years ago when some just appeared to fall apart in the sky, and they ended discovering micro fractures from the edges of the window frame caused the roof to peel off in flight. Carrier landing aircraft have airframes with much more severe time limits on them. Some 747s flying today have amazing hours clocked up, carrier aircraft last a fraction of the time.

To use my system as ground brake for landing planes, is a lot safer and easier than wheel brakes of the planes. To see how easy it is use a barrel to move a heavy load. If you push the barrel you will have to use a lot of muscle to push it. But if you pull the barrel you will feel it is a lot easier.

In the case of my system, the flat ropes, which you call cables, released by landing planes automatically and go back to their previous position automatically. The actual act of changing, replacing, the flat ropes will be done by machines within a minute or so.
 
Upvote 0

Swisaw

Free Member
Sep 24, 2010
1,849
149
London
It has nothing to do with the plane rolling over it. The metal rope is stretched until it loses its tensile strength and breaks.

Honestly, as has already been said Swisaw, the US military has a budget of over $600,000,000,000 (600 Billion! :rolleyes:) every YEAR, if there was a material on the market which had the right properties and better performance, they'd already be using it.

Yesterday on tv I saw a US fighter plane landing on a carrier with a short landing runway. The plane brought under control by an arrestor hook at the back of the plane and a cable on the runway.

It is quite obvious the US Navy System has many disadvantages. For the start, the cable is lying on the ground and the hook dragged on the ground behind the plane. So the probability of the hook missing the cable is high. The hook drags the cable on the ground. This also has a lot of quick wears and tear.

In the case of my system, the cable in not lying on the ground, It is hanging above the ground, which has the advantage of that the hook will never miss it and both the hook and the cable doesn't need to be dragged on the ground.
 
Last edited:
Upvote 0
It is quite obvious the US Navy System has many disadvantages. For the start, the cable is lying on the ground and the hook dragged on the ground behind the plane. So the probability of the hook missing the cable is high. The hook drags the cable on the ground. This also has a lot of quick wears and tear.

I'm afraid that, once again, you have it completely wrong...

The cable is held off the deck by leaf springs as shown in this picture:

http://en.wikipedia.org/wiki/File:Aircraft_Carrier_3-wire.jpg

John
 
Upvote 0
The hook drags the cable on the ground. This also has a lot of quick wears and tear.

The hook dragging the cable on the ground imparts little actual wear - certainly nowhere near the amount of wear that having a 14 tonne aircraft (F18 Super Hornet) trying to stretch the cable every time is lands has...

John
 
Upvote 0

paulears

Free Member
Jan 7, 2015
5,653
1,661
Suffolk - UK
USAF and indeed RAF airfields that have emergency diversion status have numerous systems for stopping aircraft with failed brakes, failed landing gear, and all sorts of gizmos like pop up webbing fences at the end of the runways - but NONE have arrestor cables, even when the aircraft they serve have tail-hooks. High arrestor wires would be rather dangerous anyway - imagine them going up OVER the tyres instead of the tyres going over them!

On carriers, the risk of missing the cable is VERY high, so they have 3, but you've also forgotten that the aircraft land with power ON - they don't coast in, because jet engines do not spool up instantly, so on a short field, missing the wires means FULL power instantly is required. Passenger aircraft do not land at FULL power (and the resultant speed), they land as slowly as they can while retaining control, and the long runway is needed to allow them to get off again. The power on landings on carriers are what mean MASSIVE strength is required.
 
Last edited by a moderator:
Upvote 0

Swisaw

Free Member
Sep 24, 2010
1,849
149
London
I'm afraid that, once again, you have it completely wrong...

The cable is held off the deck by leaf springs as shown in this picture:

http://en.wikipedia.org/wiki/File:Aircraft_Carrier_3-wire.jpg

John

Thanks for the picture.

I didn't see it like that on TV. I saw the cable clearly lying on the ground. In that case the hook shouldn't miss. The reason for missing is that the hook is not fixed and can jump up and down as dragged towards the cable. By the time reaching the cable it may jump upwards. This is a serious disadvantage. Add to that the cable still dragged on the ground.
 
Upvote 0
The reason for missing is that the hook is not fixed and can jump up and down as dragged towards the cable. By the time reaching the cable it may jump upwards. This is a serious disadvantage. Add to that the cable still dragged on the ground.

Off course the hook isn't fixed, it couldn't possible be and still work :rolleyes:

John
 
Upvote 0
USAF and indeed RAF airfields that have emergency diversion status have numerous systems for stopping aircraft with failed brakes, failed landing gear, and all sorts of gizmos like pop up webbing fences at the end of the runways - but NONE have arrestor cables

I'm afraid that they do - almost all US military airfields have permanent arrestor gear and they also have 'temporary' systems that can be deployed to other airbases.

Here's a pic of an F16 (which isn't even a carrier based aircraft) caught on the wires at an airbase in Iraq:

http://en.wikipedia.org/wiki/File:F-16_field_arrestment.jpg

John
 
Upvote 0

Subbynet

Free Member
Aug 1, 2005
6,000
1,101
44
Luton
Yes as John says, they don't use it daily on (land) based airfields because there is little reason too, but if required, for example they can't set-up a full size landing strip, they have a system called Portable Aircraft Arrestor Gear. Which is basically the same as the ship version but they anchor it into the ground.

Admittedly it is hard to research and find information about these systems on the internet, but here's a couple of references.

http://www.raf.mod.uk/rafwittering/aboutus/whoisbasedhere.cfm

http://www.boeing.com/assets/pdf/commercial/airports/faqs/arrestingsystems.pdf
 
Upvote 0

Swisaw

Free Member
Sep 24, 2010
1,849
149
London
I'm afraid that, once again, you have it completely wrong...

The cable is held off the deck by leaf springs as shown in this picture:

http://en.wikipedia.org/wiki/File:Aircraft_Carrier_3-wire.jpg

John

look at this one, the cable lying on the ground without spring leaves. You can also see the hook behind a landing aircraft. There are two more cables at the front, also lying on the ground.

http://upload.wikimedia.org/wikipedia/commons/6/66/US_Navy_020312-N-7265D-005_F-14_.jpg
 
Last edited:
Upvote 0

Subbynet

Free Member
Aug 1, 2005
6,000
1,101
44
Luton
look at this one, the cable lying on the ground without spring leaves. You can also see the hook behind a landing aircraft. There are two more cables at the front, also lying on the ground.

http://upload.wikimedia.org/wikipedia/commons/6/66/US_Navy_020312-N-7265D-005_F-14_.jpg

There are normally 3 or 4 cables in case the plane misses it. Having more cables means a higher chance of catching one.

You can see the leaf springs in the photo keeping the cable a few inches off the deck.

Leaf-Springs.png
 
Upvote 0

Swisaw

Free Member
Sep 24, 2010
1,849
149
London
The hook dragging the cable on the ground imparts little actual wear - certainly nowhere near the amount of wear that having a 14 tonne aircraft (F18 Super Hornet) trying to stretch the cable every time is lands has...

John


It is a question of comparative cost. A cable to arrest a 14 tonne aircraft has to be a lot stronger compared to a cable made to stop a fighter plane.
 
Upvote 0

Swisaw

Free Member
Sep 24, 2010
1,849
149
London
There are normally 3 or 4 cables in case the plane misses it. Having more cables means a higher chance of catching one.

You can see the leaf springs in the photo keeping the cable a few inches off the deck.

Leaf-Springs.png

Many thanks. The springs help to hook to catch the cable but the end of the cables fixed directly to the ground, which means the cables dragged on the ground a fair amount.
 
Upvote 0

Subbynet

Free Member
Aug 1, 2005
6,000
1,101
44
Luton
It is a question of comparative cost. A cable to arrest a 14 tonne aircraft has to be a lot stronger compared to a cable made to stop a fighter plane.

A fighter plane weights around 25-30 tonnes.

Aircraft at passenger airports weigh anything from 60-70 tonnes, all the way up to around 600+ tonnes. (In the case of the A380 - the largest passenger plane)

So a massive difference.
 
Upvote 0

fisicx

Moderator
Sep 12, 2006
46,668
8
15,360
Aldershot
www.aerin.co.uk
Many thanks. The springs help to hook to catch the cable but the end of the cables fixed directly to the ground, which means the cables dragged on the ground a fair amount.
The cable isn't fixed to the ground. There is a stonking great hydraulic system mounted under the flight deck.

When I went onto the Nimmitz (a few years back) there was a team of about 50 just looking after the system. It's not automated, you will need a crew resetting the system after each landing.

http://commons.wikimedia.org/wiki/F...the_arresting_wire_is_caught_by_the_tailh.jpg
 
Last edited:
Upvote 0

Subbynet

Free Member
Aug 1, 2005
6,000
1,101
44
Luton
Apparently the cables catch the aircraft while still in the air. In the case of my system, the aircraft lands completely before reaching the cables on the way..

The plane isn't really "in the air", at most its 6 inches off the ground because the hook doesn't lower by a great amount - the hook is pretty much at the same height as the wheels.

You can see this better in the video Scalloway posted earlier than the one I posted.

The trouble is if the aircraft lands completely you roll over the cables, probably pushing them on to the ground (by knocking them off the leaf springs) and you'll have a greater chance of the hook not catching on a cable.
 
Upvote 0

Swisaw

Free Member
Sep 24, 2010
1,849
149
London
Will full military harness type seat belts have to be fitted for every passenger ?

In the case of military, the plane is still almost in the air when the hook catches the cable. This obviously cause a violent backward reaction. You can see the cable jumping upward as it is running behind the plane to catch it in the air and bring it down, similar to a wild cat hitting a flying bird and throw it on the ground.

My system doesn't work like that. The main concept of my idea is to bring a plane to a standstill quicker so that more planes can land during any time. This obviously reduces the length of the runway for the landing of each plane. But the runway still have to be long enough for a plane to land completely before reaching any cables on the way. Obviously the cables have to be set in a manner not to catch planes violently but after catching by a second or so their pulling power increases as the speed of the plane decreases. This should cause a gentle brake without any adverse affects on the passengers.
 
Last edited:
Upvote 0

Swisaw

Free Member
Sep 24, 2010
1,849
149
London
I think you mean 505,000,000 Joules for an aircraft landing

(based on your figures: KE = 1/2 m x v squared:

1/2 x 500,000 x 45 x45 = 506,000,0000

505,000,000 Joules = 140.3 KWh (I used Wolfram Alpha for this calc)

So if you spent £218.66 for 5,831.11 KWh then 140.3 KWh is worth about £5.26

I haven't read the whole thread in detail, but I get the impression that you are trying to harness the energy 'given up' in slowing a jet to a stop on landing. If the figures above are correct, given a near 100% efficiency (nigh on impossible) you are only going to generate the equivalent of £5 worth of gas for each jet that lands.

I must confess, the figures seem lower than I might have thought, there might be an error in there somewhere.

J

Many thanks Jeff.

This morning I managed to it by myself. I got £5.27.

Airports using my system should become self-sufficient in energy supply. This should save more than enough to run and maintain the system. This is in addition to the green and cost reduction benefits. The airport needs less runways, cost, and receives more aeroplanes, revenues.
 
Upvote 0

Latest Articles

Join UK Business Forums for free business advice