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Home > Airways > Comparison of the Airbus A320's speed...
Comparison of the Airbus A320's speed...
Flying - Airways
Monday, 12 February 2007 22:02

Dear Captain Lim,

As you know I went overseas for my usual year-end holiday, this time to China. I flew in the plane you are currently piloting. Instead of sitting quietly like a sleeping Buddha like all other passengers usually do, I spent my time recording whatever data I could get hold of during he flight with just only a watch, a compass, pen and note book, and later back home in Malaysia, a scientific calculator. After that, came some crazy food for thought below. As through our telephone conversation following my previous e-mail to you, and while waiting for you to give me the answers to my questions while you shift house, here are some observations I made from my flight. This follows some mad thoughts which I am now send to you.

I suppose I have a compulsive obsession disorder all my life to observe, measure, analyze, and to conclude as a trained research scientist.

December 16, 2006

Plane from LCCT to Macau was a A320 (180 seats) airbus

Plane started to accelerate along the LCCT (KLIA) runway at 4:24:00 p.m.

Plane lifted off at 4:24:44 p.m.

Took 44 seconds to lift off runway

Plane landed in Macau Airport at 7:45:33 p.m.

Plane stopped completely at 7:50:43 p.m.

Time taken from KLIA to Macau: 3 hrs 20 min 49 sec (3.346944444 hours)

Time taken to taxi to a complete stop: 5 min 10 sec

Just before departure, the plane was slowly taxing into the runway. At the beginning of the runway, the plane stopped taxing briefly for some 30 seconds before the engines suddenly roared to accelerate the plane along the runway until the wheels just lifted off. The time of the acceleration was recorded as 44 seconds at KLIA, and on the return flight to Malaysia, it was 36 seconds at Macau International Airport.

But I do not know the distance it had traveled along the runway or the final speed just before lifting off the runways. There was no means for me to measure this ? wished I could. But the initial speed was 0 meter per second as the pilot briefly waited for about 30 seconds for instructions from the control towers before accelerating.

December 23, 2006

Plane from Macau to LCCT was an A320 airbus

Plane started to move at 8:29:30 p.m.

Plane accelerated at 8:36:18 p.m.

Plane lifted off runway at 8:36:54 p.m.

Therefore acceleration lasted for 36 sec

Plane touched LCCT runway at 11:59:49 p.m.

Therefore plane took 3 hrs 22 min 55 sec from Macau to KLIA (runway-to-runway)

Plane taxied to a complete stop at 12:05:54 a.m. on December 24, 2006

Therefore time taken to taxi to a complete stop after touch down was 6 min 5 sec

Distance and Speed announced by the Pilot:

However, on the return flight from Macau when plane was near KL, the pilot announced it was 280 km away from KLIA, and it will take about 20 minutes before landing.

With that information I quickly jotted down, the estimated average speed of plane must have been:

14 km per min or 840 km per hour.

Theoretical Distance and Velocity

However, do not know the actual distance between the airports of Macau and Kuala Lumpur, except it was a theoretical 2481.57 km along the Great Circle as calculated out by using the Haversine formula and / or by applying the spherical law of cosines ? based on spherical trigonometry, and the geographical coordinates (spherical coordinates) of these two points on Earth? This is the shortest possible distance in theory. Probably it is must further than this. A plane needs to ascend up to 10 km above the surface of Earth, and then descend again, all at a gentle angle, besides probably having to deviate the flight path slightly on the way? There is no way of determining distance by time taken alone, or multiplying the estimated velocity x time, since we need to know the exact velocity, and that velocity has to be exactly the same (stable) from runway to runway. This is not possible since may speed up from a tail wind, or slowed down by a head wind. It also has to reduce its speed just before landing or at the instruction from the control tower. So I really do not know the exact practical distance between the two runways. Only a pilot with his sophisticated instruments and computers in the cock pit can tell us.

If only I know the landmarks below:

However, when I flew I could see the land below and the wing and engine on my side sweeping over the land below. If only I knew where I was, I could use the tip of the wing or some reference point(s) on other parts of the wing or engine as an indicator to record the time taken to sweep two points on the ground, and if their distances are known, I could easily estimate the velocity of the plane at that point in time. Then by taking many, many similar readings, we could easily calculate out the average velocity of the plane, and their directions from my pocket compass. From these we can easily plot out *home-made* flight path, and distance, calculate out the velocity from time taken by using a good map, a protector (to measure out the degrees and direction the plane took), a divider (to mark out the time and distance covered), compass (to mark out the bearing), and finally determine the speed and total distance covered when we get back home. But at last, even this simple *luxury* I do not have as a passenger ? at least not with A320 Airbus.

On board flight data with SIA and Qantas:

When I traveled from KL to London or to Australia in Jumbo jets with Singapore Airlines, or with Qantas on several occasions, there was always an onboard screen in the cabin to screen out the flight data ? velocity, flight path, time left the last airport, time remaining to arrive the next airport, the temperature on the ground, the temperature outside (usually around ? 55 degrees Celsius), velocity of head wind, tail wind, etc.

How exciting and entertaining these data are to a scientist like me, albeit a scientist engaged in medical research, and not a physicist specializing in aviation physics. I enjoy observing, recording, and analyzing the data later, while other passengers slept all through their boring flight, or watched or entertained by those stupid in-cabin fighting movies when they wake up. I always stayed awake, and recorded all these data at every moment of the flight into my note-book each time I flew in Jumbo jets, but I could not get these scientific luxuries with the Airbus A320

Flight Path

The only thing I can record was the time taken from my stop watch, and the direction of flight from the fairly good pocket compass I brought along with me. It showed a very stable path in the SW direction (Macau-KL). That was the only *navigational aid* I had on board. I wanted to bring a barometer and a hygrometer along to measure the air-pressure and humidity in the cabin and outside later, but it was troublesome. I was on a holiday, not there to conduct some scientific experiments. But I brought a sensitive scientific thermometer with me which recorded the temperature at Guilin at 8 deg. Celsius one day at its coldest.

Since I do not know the speed of the plane, its change of speed due to head and tail winds, or instruction from the Control Tower, by how much, and over what distances, I can only obtain the distance by calculations in theory, working on some assumptions. This was of course the shortest possible theoretical distance at sea-level. At 10 km up there of course the real distance would be much more.

I wish I have all these information which only instruments can measure. I was only a passenger in the plane without any navigation instruments, except a small hand compass and a stop-watch to merely to show the direction of the flight path and time taken. The compass showed a stable 215 deg (NW) azimuth from Macau to Kuala Lumpur during the 3 hrs 22 min 55 sec flight.

Since the recorded flight time from KLIA to Macau Airport runways was 3 hrs 22 min 55 sec (3.381944 hrs), the velocity (speed with direction) of the plane must have been at least 733.77 km per hour based on a minimum theoretical distance of 2481.57 km (1541.97 statute miles) between the runways of KLIA and Macau IA I pre-determined earlier. That*s the best I could derive theoretically. In practice, it must have been much more than this?

Direction of Flight

During the flight from Macau to KL, my compass measured the following:

Azimuth 205 deg (SW) soon after take off
195 deg azimuth about an hour after leaving Macau
220 deg azimuth at 10:39 p.m.
210 deg azimuth at 11:29 p.m. (30 minutes before landing)

I could have taken much more reading than this, if not for my wife, and other passengers talking to me most of the time ? which was a nuisance to me. Anyway, the few data over time gave some guidelines and indications that the flight path was very straight and stable all the way.

Even just on landing, the compass still showed about the same direction and orientation of the LCCT runway as during the flight. The flight path was almost very consistent in a south-westerly direction all through the 3 hrs 22 minutes of flight. It was so unlike in the tour bus in China where the bus kept changing directions especially in the cities and towns, and the compass had to change bearing accordingly. Understandable!

Here*s some info I got about the A320 Airbus:

Cruising speed 0.79 Mach = 967.75 kph

Maximum speed 0.82 Mach = 1004.5 kph

Based on a temperature of 15 degrees Celsius, Mach 1 is 340.3 m?s-1 (1,225 km?h-1 or 761.2 mph) in the atmosphere.

Using the above info on the speed of the plane, here is a simple reduction about the distance, assuming the speed was constant (it was not):

If A 320 Boeing took 3 hrs 20 min 49 sec (3.346944444 hrs) to fly from KLIA to Macau International Airport, and if the cruising speed was 967.75 kph (based on the above info), then KLIA to Macau IA must have been 3239 km (at cruising speed), or 3362 km (at max speed). Of course again this is a very rough estimate because we assume the plane maintained a constant speed.

Compare this figure of 2481.57 km (1541.97 miles) obtained by theoretical calculations between KLIA and Macau International Airport using their geographical coordinates into the spherical trigonometry equations. Naturally the theoretical distance would be much shorter because it took the shortest possible distance at ground level instead of high up at 10 km altitude of a plane. A plane will have to circumnavigate a greater distance over the Great Circle (circumference = 2pr) as its radius from the centre of the Earth is an additional 10,000 metres, let alone the additional distance needed for the ascend, and the descend of the aircraft.

Velocity of space crafts:

Now let us compare this speed with that of a space-craft. We can reduce this by calculation from some data taken from NASA Astronaut Dr. Edward Tsang Lu, PhD website:

According to the website, Dr. Lu traveled 3.6 million miles in 144 orbits of the Earth logging a total of 9 days, 5 hours, 19 minutes, and 55 seconds in space.

If that was the case, the velocity of Dr Edward Tu spacecraft in a circular orbit must have been:

4.518 miles / sec = 7.271 km / sec
= 271.086 miles per min = 436.27 km / min
= 16265.16 miles per hour = 26176.24 km / hr

Hence a spacecraft would be:

26.059 times faster than an A320 Airbus at max speed
27.048 times faster at cruising speed

Actually I suspect the above info that Dr Edward T Lu gave in his website as ?3.6 million miles in 144 orbits of the Earth logging a total of 9 days, 5 hours, 19 minutes, and 55 seconds in space? was a little too conservative. This is because when we divide 3.6 million miles by 144 orbits, we get the circumference of each orbit. This would only be 25,000 miles (40234 km) each orbit, which is almost the same circumference of the world on the surface (sea-level) along the equator.

Space crafts are high up:

But space craft and satellite don*t orbit the Earth at sea-levels. Since the equatorial radius (r) is 6378.137 km (3963.19 statute miles), the circumference 2pr = 40075 km (24900 miles) which is almost the same as what Dr Lu given in his website, albeit rounded up to the nearest figure.

But a space craft is high up over the surface of the Earth. Various satellites and spacecrafts orbit at different altitudes. A geostationary satellite for instance, orbits the earth directly over the equator at approximately 22,240 miles (35.790 km) up. Such an orbit is called a geosynchronous orbit (GEO). Over this altitude, one complete trip around the earth (relative to the sun) takes 24 hours. Thus, the satellite remains over the same spot on the earth*s surface at all times, and stays fixed in the sky from any point on the surface from which it can be "seen." So-called weather satellites are usually of this type.

So, if we divide 9 days, 5 hours, 19 minutes, and 55 seconds by 144 orbits, we deduced that each orbit took 92.22 minutes, and not 24 hours to stay at one spot relative to Earth*s rotation. This means that Dr Tu was never on a geostationary satellite. Such a spacecraft is only possible if it was placed at an orbit at least 22,240 miles (35.790 km) away from Earth according to Kepler*s Third Law for the period of rotation.

Although Kepler*s Third Laws normally applies for planets around the Sun, the same law applies for satellites too. The orbit Dr Tu has taken has to be at a much lower altitude. But he did not tell us at what altitude, or in which space station or spacecraft he was in. So we cannot be too certain except make some guesses.

Which Space Station?

The Russian space station Mir for instance dropped from its original altitude of about 195 miles (315 kilometers) to an altitude of about 143 miles (230 km). Mir is now in a slow descent. But the International Space Station (ISS) orbits our planet about 400 km high.

Let us presume Dr Lu and other international astronauts were inside the ISS as a NASA astronaut. If that was the case, the radius of ISS from the centre of the Earth would be:

Equatorial radius of the Earth + altitude of ISS

This means the circumference of a single orbit should be 2p (6378.137 + 400) km = 42588 km. Since he made 144 orbits, the total distance covered in 9 days, 5 hours, 19 minutes, and 55 seconds (13279.916 minutes) would be 6132,672 km.

Hence the recalculated speed of his spacecraft or space station ought to be (presumed that 13279.916 minutes was correct):

6132672 km ? 13279.9 min =

4.78 miles / sec = 7.69 km / sec
286.9 miles / min = 461.8 km / min
17217 miles / hr = 27708 km / hr

(If we presumed that 13279.916 minutes in orbit was correct)

And, not:

4.518 miles / sec = 7.271 km / sec
271.086 miles per min = 436.27 km / min
16265.16 miles per hour = 26176.24 km / hr

(As it was deduced from incomplete info given in the website). But I did not use my recalculated data based on his. I trusted Dr Edward Lu figures. He is an astronaut, and I am not. There is no comparison.

Let*s now look at other speeds:

Normal average walking speed is about 4.95 feet /sec (1.5 metres / sec) = 5.43 km per hour
Normal average speed of a bicycle = 12 mph (19 km / hr)
A car average speed is 62 mph = 100 km / hr
Assuming the average cruising speed of the A320 Airbus I was in, was a little lower at 950 km / hr

Then we have:

A man riding a bicycle at normal speed is 3.5 times faster than a pedestrian
A car at 100 km / hr is 5.3 times faster than a cyclist at average speed
A car driven at 100 km / hrs is 18.4 times faster than someone walking along the road.

An Airbus A320 cruising at 950 km / hr. is:

9.5 times faster than a car at 100 km / hr
50 times faster than a cyclist
175 times faster than someone walking on the ground at 5.43 km

But a spacecraft hurtling in space at 26176.24 km / hr is:

27.6 times faster than the Boeing Airbus A320 I was on
261.8 times faster than the car my son was driving me home from the airport
1378 times faster than when I cycled a few days later along the lonely old road from Gombak towards Genting Highland for my evening exercise
And finally, 4820 times faster than when I walk to a nearby market.

The Ultimate of All Speeds:

But at last, even more amazing is a speed which God created it Himself ? the speed of light at:

299792458 m /s = nearly 300,000 km / sec = 18 million km /min = 1080 million km / hr

Let*s use the speed of light in a vacuum as a yardstick. This means that light is faster than the following by:

41,259 (41.3 thousand) times faster than a space-craft
1136,842 (1137 thousand) times faster than an Airbus A320
10,800,000 (11 million) times than a car at 100 km / hr
56,842,105 (57 million) times than a bicycle at 19 km / hr
198,895,028 (200 million) times faster than someone walking at 5.43 km / hr.

But even more amazing is one of the shortest distances, to the ultimate of all distances:

These are, the width of the hydrogen atom that permeates the entire universe to the diameter of the universe itself. The diameter across the width of the hydrogen atom is taken as the smallest unit reference of length in the entire universe since the hydrogen atom is the symbol of the commonest element in the entire Universe.

The Data:

The diameter of a single proton is 10-15 metres
The diameter of a hydrogen atom is 10-10 metres
The diameter of the universe by the latest estimates at least 30 or 40 billion light years in diameter.

This means that light traveling at a velocity of 299792458 metres per sec will take only:

3.34 x 10-18 seconds (3.34 quintillionth of a second) to cross the diameter of a hydrogen atom at ground state.

3.34 x 10-23 seconds (33.4 septillionth of one second) to cross the diameter of a single proton which is one of the smallest atomic particles.

But light will have to travel for:

299792458 metres ? 1000 (to change into km / sec) x 60 sec (to change to km per min) x 60 min (to change to km per hour) x 24 hr in a day x 365.25 days in a year x 40,000, 000,000 years = 3.78 x 1023 km (378 sextillion km) in order to cross from one end of the universe to the other. This will take light 4 x 1010 (40 billion) years to achieve that! Fancy that!

The Ultimate of Time:

Probably the shortest time possible, is to use the diameter across the width of a hydrogen atom as the smallest distance, and the speed of light in a vacuum as the fastest of all speeds as reference yardsticks to measure time. From the above this would be 3.34 x 10-18 seconds (3.34 quintillionth of a second).

The Ultimate of Temperature:

In similar light, the ultimate of all theoretical possible temperatures can only be:

V = 0.158 v (T ? m).

By using a series of equations, it can also be shown that the above equation can also be expressed as:

T = 40 mv2

The factor 40 only holds if we use units of temperature in (degrees) Kelvin, and velocity in km / sec.

Where, V = velocity of any atomic particles in km / sec (molecular / atomic velocity)

T = temperature in degrees Kelvin

M = mass of particles at rest

Let us set the value of *v* (velocity of the molecules as a gas gets heated up) as the maximum possible speed of 299, 792 km / sec. ? the speed of light. Applying this value into the equations, we will get what seems the maximum possible temperature.

When the temperature reaches higher and higher, the atomic particles move faster and faster until it reaches the velocity of light. Since nothing can travel faster than light, then the highest possible theoretical temperature can only be:

T max = 3.59 x 1012 = 3,600,000,000,000 (degrees) Kelvin (3.6 million, million K)

At last! It is not as simple as that! At searing temperatures of millions of degrees all molecules and atoms break down into mere particles. Fusion reactions between simple nuclei are possible so that complicated nuclei can be created. At even still higher temperatures (if this is theoretical possible), then the reverse is true, and all nuclei must break down into even simpler particles. This gives rise to further complications between creation / destruction of matter, temperature, pressure, dimension and time.

Mass Increase with Velocity

But by then, as the velocity increase to near that of light, the mass of the particles will also increase according to Fitzgerald-Lorentz contraction derived from Einstein Theory

M = m 0 / m ? [v (1- v2 / c2)]

Where M= final mass, m 0 is the initial rest mass, v2 = velocity of the particle at rest, c2 = the velocity of light.

As the particles go faster and faster their mass get more and more massive, until their mass reach infinity at the speed of light. So when v = c, the equation tells us the final mass (M) becomes infinite - exceeds that of the mass of the Universe itself. That is not possible. This means that even the temperature of 3.6 million, million Kelvin can never be reached.

But let me not go further into particle and plasma physics as I am not any where near a particle and high temperature physicist, else the real experts may hammer me to death for trying to be a *ka-po-chee* (busybody).

That*s Light Fantastic!

How does an Airbus A320 compared to that?

With that, how long would it take to?

walk / bicycle / go by car / fly by A320 Airbus / and finally whiz across in a space-craft over these following dimensions?

The diameter of a proton
The diameter of a hydrogen atom
The diameter of the Universe

Well, I am not going to tell you the answers as I am going to have my fish head curry dinner now. My wife has prepared the fish-head for me since this morning, and I don*t think that a fish head has much brain to think about this problem.

So it is better that I leave this to you with your larger head to sort this out for me. These 3 questions will keep your heart and brain young forever.

But for the moment, this is the simple joy I got in this story from my simple flight from KL to Macau and back.



Dr JB Lim

Hi Dr JB Lim,

Thank you for your very detailed (& lengthy!) description of your flight to Macau and the answers to your own queries. You are certainly a very exceptional air traveler that attempts to calculate and compare the speed of the plane to other mode of transport or motion to such a great precision! Perhaps this is one way to arouse your scientific mind and keep your heart and brain young forever :-)!

Well, I know it is not easy to compute any speeds without the proper instruments but your answers, especially dealing with the Airbus flight were generally accurate. Yes, the speed of the A320 varies with the winds of the day and any restrictions as imposed by the air traffic controllers.

As regards to how long it would take to walk, cycle, go by car, fly an Airbus A320 and finally to whiz across in a space-craft over:-

a. The diameter of a proton
b. The diameter of a hydrogen atom
c. The diameter of the Universe,

I am afraid my simple mind is not able to give you the answers that you have already done. Yes, that is left to the realm of your scientific mind and you have done a great job!

Best regards,

KH Lim.


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With so much interest in collecting flight data and performing scientific calculations. I would like to suggest Dr. Lim to buy a handheld GPS receiver. In US the range starts from around 400$ but I am not sure which features you'll need for your purpose. And also the legality of whether the passengers are permitted to carry a GPS on board.
Sincere thanks and respect to Capt. Lim for such an informative website.
Vishwas Gaikwad , 14 Feb, 2008

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