B737NG retractable landing light drag.
Thread Starter
B737NG retractable landing light drag.
I'm trying to calculate the extra drag created while the retractable landing lights are extended.
Ultimately I'm wanting to calculate the extra work and hence energy wasted by having these extended below 10,000 ft on descent.
I have found a website that can calculate force on various shapes given certain variables:
Drag Coefficient - DiracDelta Science & Engineering Encyclopedia
I'm assuming that the lights are hemisphere in shape and have a diameter of 0.27 m.
To keep things simple (and a little rough) I entered air density as ISA SL (1.225 kg/m^3)
Cd = 1.17
At 250 kt (125 m/s) it calculated the drag as 2564 Newtons.
2564/9.8= 261 kg.
Is that correct? It seems like a lot. I did a confidence check by entering the formula with 53 kt (100 km/h) and it came up with 119 N (12 kg) which seems reasonable if you imagine having such a shape out the car window and feeling the drag with your arm extended.
Anyway I was then wanting to compare the extra drag with total drag to convert it to a percentage of total drag.
If the aircraft has a L/D ratio of approx. 20:1 and the aircraft has a gross weight of say 65 tons (Max Landing Weight) the total drag would be approx. 3250 kg. So as a percentage, a single landing light extended at 250 kt would increase the drag by 8%. (261/3250). That doesn't sound right to me. I think it's out by a factor of ten but I can't see where the calculations go wrong.
I know the MEL increases the fuel burn by 1% per extended light so 0.8% sounds more like it.
And assuming a straight line descent from 10,000 ft to touchdown the distance travelled would be approximately 30 nm or 55.5 km (55,500 m)
Total energy in Joules is Newtons multiplied by distance (m). So according to the numbers above the amount of energy wasted is:
2564 N x 55,500 m = 142,302,000 Joules (142 MJ)
Can someone please confirm these numbers? Thank you.
Ultimately I'm wanting to calculate the extra work and hence energy wasted by having these extended below 10,000 ft on descent.
I have found a website that can calculate force on various shapes given certain variables:
Drag Coefficient - DiracDelta Science & Engineering Encyclopedia
I'm assuming that the lights are hemisphere in shape and have a diameter of 0.27 m.
To keep things simple (and a little rough) I entered air density as ISA SL (1.225 kg/m^3)
Cd = 1.17
At 250 kt (125 m/s) it calculated the drag as 2564 Newtons.
2564/9.8= 261 kg.
Is that correct? It seems like a lot. I did a confidence check by entering the formula with 53 kt (100 km/h) and it came up with 119 N (12 kg) which seems reasonable if you imagine having such a shape out the car window and feeling the drag with your arm extended.
Anyway I was then wanting to compare the extra drag with total drag to convert it to a percentage of total drag.
If the aircraft has a L/D ratio of approx. 20:1 and the aircraft has a gross weight of say 65 tons (Max Landing Weight) the total drag would be approx. 3250 kg. So as a percentage, a single landing light extended at 250 kt would increase the drag by 8%. (261/3250). That doesn't sound right to me. I think it's out by a factor of ten but I can't see where the calculations go wrong.
I know the MEL increases the fuel burn by 1% per extended light so 0.8% sounds more like it.
And assuming a straight line descent from 10,000 ft to touchdown the distance travelled would be approximately 30 nm or 55.5 km (55,500 m)
Total energy in Joules is Newtons multiplied by distance (m). So according to the numbers above the amount of energy wasted is:
2564 N x 55,500 m = 142,302,000 Joules (142 MJ)
Can someone please confirm these numbers? Thank you.
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I'm not wishing to dampen your enthusiasm and endeavour, but why would you have them extended below FL100? They are tiny things. You have a big a/c with TCAS, big fixed landing lights, probably radar but AUS may have much more class E airspace than EU. What extra use do they give you in 'see & be seen'?
Boeing recommend (I think) them retracted after takeoff both for drag and stress on the fixtures at 250kts.
Boeing recommend (I think) them retracted after takeoff both for drag and stress on the fixtures at 250kts.
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Boeing has got rid of them all together on the new B737 Max.
OMG. New type rating needed. Some airlines will need to pay for a new "cleared for takeoff & landing" switch.
OMG. New type rating needed. Some airlines will need to pay for a new "cleared for takeoff & landing" switch.
Thread Starter
RAT 5
Indeed! Note in my original post I said I was wanting to quantify the energy WASTED by having the lights extended.
At the moment it is standard company procedure to have them extended below 10,000 ft regardless of airspeed, and yes it is for anticollision purposes! I am hoping to have some influence to change this policy by pointing out how much energy and hence fuel is being wasted on a fleet-wide basis every day.
I need the collective knowledge of pprune to help me with this endeavor.
Indeed! Note in my original post I said I was wanting to quantify the energy WASTED by having the lights extended.
At the moment it is standard company procedure to have them extended below 10,000 ft regardless of airspeed, and yes it is for anticollision purposes! I am hoping to have some influence to change this policy by pointing out how much energy and hence fuel is being wasted on a fleet-wide basis every day.
I need the collective knowledge of pprune to help me with this endeavor.
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I'm assuming that the lights are hemisphere in shape and have a diameter of 0.27 m.
Also, is the pressure at the point where the lights are located the same as elsewhere on the fuselage? Do parts of the fuselage create a protective bow-wave or is this offset by AOA.
Blip,
I got the same number as you did by using .27 as the radius of the light. If it's really .27 M in diameter, you need to divide by 2 to get the radius, then square and multiply by Pi to get the area. (Or use the square of the diameter and divide by 4 at the end.) Assuming the mistake wasn't in your post, the value you got is 4 times too high. (Of course, that's per light.)
I got the same number as you did by using .27 as the radius of the light. If it's really .27 M in diameter, you need to divide by 2 to get the radius, then square and multiply by Pi to get the area. (Or use the square of the diameter and divide by 4 at the end.) Assuming the mistake wasn't in your post, the value you got is 4 times too high. (Of course, that's per light.)
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On the Fokker 100 the wing tip mounted landing lights auto retracted in the event of an engine failure.
If a landing light was MEL'd as stuck out then the penalty was 973KG off the MTOW.
If a landing light was MEL'd as stuck out then the penalty was 973KG off the MTOW.
Oh goodness.....
Who cares really?
If it's a problem it will be in the 737 FCOM/MEL
If not then....
You sit there calculating the cost while I workout what to order for lunch.......it's a tough choice, BCF ( Beef Chicken or Fish )
Who cares really?
If it's a problem it will be in the 737 FCOM/MEL
If not then....
You sit there calculating the cost while I workout what to order for lunch.......it's a tough choice, BCF ( Beef Chicken or Fish )
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Saved fuel burn from T/O to FL100 /10.000' is approx 12 kgs, by keeping retractable ldg lights in retracted position. As per Boeing.
New NG can be delivered with a LED LDG light installation, with no retracts, better lights, and weight saving.
New NG can be delivered with a LED LDG light installation, with no retracts, better lights, and weight saving.
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Didn't Ryanair look into this and change their procedures to retract them after T/O instead of at 10,000ft?
Believe this was a number of years ago and actually saved them quite a bit of fuel on a yearly basis.
Believe this was a number of years ago and actually saved them quite a bit of fuel on a yearly basis.
Thread Starter
Thank you Chu Chu. That's exactly what I was asking for. And thank you to all the others who contributed constructively to the discussion.
By the way I did a walk around the other day and measured the diameter as 23 cm. And yes there is a very slight concave to the lense.
ACMS Thanks for your input. I too make a meal choice then turn my mind to other things.
By the way I did a walk around the other day and measured the diameter as 23 cm. And yes there is a very slight concave to the lense.
ACMS Thanks for your input. I too make a meal choice then turn my mind to other things.
Gender Faculty Specialist
How long does it take you to choose between three dishes?
Thread Starter
OK I've done some simple calculations which reveal the region of numbers we're talking about here.
CD = 1.17
Diameter = 0.23 m
TAS = 250 kt
Force = 465 N
Distance = 55,000 m (30 nm)
Energy = 25,575,000 Joules per light.
How high would this energy lift a B737 at 65,000 kg?
25,575,000 / 9.8 / 65,000 = 40 metres (131 ft) x 2 = 262 ft
If it takes 7 mins to descend 10,000 ft....
Extra rate of descent = 262 / 7 = 37 ft per min.
CD = 1.17
Diameter = 0.23 m
TAS = 250 kt
Force = 465 N
Distance = 55,000 m (30 nm)
Energy = 25,575,000 Joules per light.
How high would this energy lift a B737 at 65,000 kg?
25,575,000 / 9.8 / 65,000 = 40 metres (131 ft) x 2 = 262 ft
If it takes 7 mins to descend 10,000 ft....
Extra rate of descent = 262 / 7 = 37 ft per min.
Thread Starter
Take off performance:
Drag at 160 kt = 190 N
10 nm = 18520 m
Energy = 3,518,800 Joules per light.
3,518,800 J / 9.8 / 75,000 kg = 4.79 m per light.
2 x 4.79 x 3.281 = 17 ft
So after 10 nm the aircraft would be 17 ft lower than it would have been if the lights were retracted.
Rough numbers I know but it give you an idea.
Drag at 160 kt = 190 N
10 nm = 18520 m
Energy = 3,518,800 Joules per light.
3,518,800 J / 9.8 / 75,000 kg = 4.79 m per light.
2 x 4.79 x 3.281 = 17 ft
So after 10 nm the aircraft would be 17 ft lower than it would have been if the lights were retracted.
Rough numbers I know but it give you an idea.