Why heavier aircrafts take longer to slow down in the air?
Thread Starter
Join Date: Apr 2008
Location: Test
Age: 35
Posts: 233
Likes: 0
Received 0 Likes
on
0 Posts
Thanks everyone for the replies.
Ok i understand that a heavier aircraft has more energy but how does it take a longer time to slow down?
Is there a better analogy?
If i were to compare a lorry to a car, you would say the lorry will coast further than a car. But can i not argue that gravity will pull the heavier lorry down more than the car and effectively slowing down the lorry faster?
Can anyone counter this? Pardon me, i don't possess any physics knowledge, just logic.
Pls advise.
Cheers
Ok i understand that a heavier aircraft has more energy but how does it take a longer time to slow down?
Is there a better analogy?
If i were to compare a lorry to a car, you would say the lorry will coast further than a car. But can i not argue that gravity will pull the heavier lorry down more than the car and effectively slowing down the lorry faster?
Can anyone counter this? Pardon me, i don't possess any physics knowledge, just logic.
Pls advise.
Cheers
Join Date: Jul 2009
Location: France - mostly
Age: 84
Posts: 1,682
Likes: 0
Received 0 Likes
on
0 Posts
Originally Posted by extricate
But can i not argue that gravity will pull the heavier lorry down more than the car and effectively slowing down the lorry faster?
But your original question was about aircraft, not cars and lorries, and the physics are somewhat different.
Last edited by HazelNuts39; 22nd Nov 2012 at 16:12.
Join Date: Aug 2007
Location: Derby UK
Age: 59
Posts: 112
Likes: 0
Received 0 Likes
on
0 Posts
simple analogy.
i throw a ping pong ball at your head at 20 MPH.
then i throw a snooker / Billiards / pool ball at the same place at the same speed (20 MPH )
which one hurts the most ??
Or if i roll a bowling ball down the alley will it knock over the pins.
now if i roll a balloon down at the same speed what happens ?????
sorry but you have to accept that the physics for now is correct and that the formula's and explanations you have are enough for you to understand why a heavier aircraft takes more slowing down than a lighter one.
Hope this helps. But if you really want to get into the physics, you really need to start at the beginning and build your knowledge up, rather than jumping in at some random point.
GB
i throw a ping pong ball at your head at 20 MPH.
then i throw a snooker / Billiards / pool ball at the same place at the same speed (20 MPH )
which one hurts the most ??
Or if i roll a bowling ball down the alley will it knock over the pins.
now if i roll a balloon down at the same speed what happens ?????
sorry but you have to accept that the physics for now is correct and that the formula's and explanations you have are enough for you to understand why a heavier aircraft takes more slowing down than a lighter one.
Hope this helps. But if you really want to get into the physics, you really need to start at the beginning and build your knowledge up, rather than jumping in at some random point.
GB
Join Date: Jul 2009
Location: France - mostly
Age: 84
Posts: 1,682
Likes: 0
Received 0 Likes
on
0 Posts
Originally Posted by Bye
you have are enough for you to understand why a heavier aircraft takes more slowing down than a lighter one.
Regards,
HN
Join Date: Aug 2007
Location: Derby UK
Age: 59
Posts: 112
Likes: 0
Received 0 Likes
on
0 Posts
I'm wondering if your simple analogy explains that the heavier aircraft slows down more rapidly than the lighter aircraft below the minimum drag speed.
http://home.anadolu.edu.tr/~mcavcar/htk224/MinDrag.pdf
GB
Last edited by Bye; 22nd Nov 2012 at 18:10.
Join Date: Jul 2009
Location: France - mostly
Age: 84
Posts: 1,682
Likes: 0
Received 0 Likes
on
0 Posts
Bye,
Post #2 states the 'basic physics'.
F/M is 'basic' aerodynamics.
Your analogy doesn't address either one.
Post #2 states the 'basic physics'.
F/M is 'basic' aerodynamics.
Your analogy doesn't address either one.
Last edited by HazelNuts39; 23rd Nov 2012 at 06:28.
Join Date: Dec 2007
Location: Montreal
Age: 92
Posts: 156
Likes: 0
Received 0 Likes
on
0 Posts
Inertia, momentum weight and mass and all that !
DaveReidUK
Join Date: Jan 2008
Location: Reading, UK
Posts: 856
Quote:
It is worth remembering that at a given weight you will always have the same inertia
Inertia is a function of mass, not weight.
On its way to the Moon, Apollo had a lot of inertia but weighed nothing at one point
If that is true, would inertia in zero gravity space be equivalent to Vsquared ?
?
It seems to me that there is not much attention paid to the cleanliness of aerodynamic profiles in modern heavies and that is also a contributing factor to it taking longer to slow down. My high performance glider, when loaded with water has a longer slowing down time when decellerating in level flight (vs a Schweitzer 222, which is a big drag queen in the gliding world !)
Join Date: Jan 2008
Location: Reading, UK
Posts: 856
Quote:
It is worth remembering that at a given weight you will always have the same inertia
Inertia is a function of mass, not weight.
On its way to the Moon, Apollo had a lot of inertia but weighed nothing at one point
If that is true, would inertia in zero gravity space be equivalent to Vsquared ?
? It seems to me that there is not much attention paid to the cleanliness of aerodynamic profiles in modern heavies and that is also a contributing factor to it taking longer to slow down. My high performance glider, when loaded with water has a longer slowing down time when decellerating in level flight (vs a Schweitzer 222, which is a big drag queen in the gliding world !)
Join Date: Aug 2007
Location: Derby UK
Age: 59
Posts: 112
Likes: 0
Received 0 Likes
on
0 Posts
post #2 states "some" basic physics but is not the correct formula to explain why a heavier object has more kinetic energy than a lighter one at a given speed and drag.
F/M is that some code to describe movies of a certain type. certainly not a formula or basic aerodynamics of any kind.
I look forwards to your explanations and analogies of why a heavier aircraft below Vmd slows down quicker than a lighter aircraft.
GB
F/M is that some code to describe movies of a certain type. certainly not a formula or basic aerodynamics of any kind.
I look forwards to your explanations and analogies of why a heavier aircraft below Vmd slows down quicker than a lighter aircraft.
GB
Join Date: Jul 2009
Location: France - mostly
Age: 84
Posts: 1,682
Likes: 0
Received 0 Likes
on
0 Posts
Originally Posted by Bye
F/M is that some code
Extricate's questions in posts #1 and #5 are about A, which equals F/M. F is aerodynamic drag, and it is related to mass M by the aerodynamic lift which equals weight. So a question about A=F/M is about aerodynamics.
I look forwards to your explanations and analogies of why a heavier aircraft below Vmd slows down quicker than a lighter aircraft.
Originally Posted by Microburst2002
The heavy one, flying at max L/D (green dot speed) would make, say 240 kt, whereas the light one would be making only 210 kt. But they are both flying at 300 kt, so the heavy one is 30 kt closer to green dot that the light one. Its L/D ratio is better. Its AoA is closer to min Drag, to green dot.
Regards,
HN
Last edited by HazelNuts39; 22nd Nov 2012 at 21:43.
Join Date: Dec 2007
Location: Pasadena
Posts: 633
Likes: 0
Received 0 Likes
on
0 Posts
If you load up the same truck with more weight, keeping the same brakes, then it takes longer to stop. If there's the same braking force, energy to dissipate scales with mass.
It's a little bit more complex with an aircraft, as if it's heavier, it needs more lift, which means a different speed, height or attitude to fly level. More lift usually means more drag, both in approximate proportion to weight, and so more power is required to cruise heavier.
When losing energy descending though, if you try to shed speed deliberately with speedbrakes, that amount of braking isn't affected by weight, just by the speed and spoiler setting, and so a heavier aircraft will need more energy throwing away, and a longer deployment. The energy required to drop from 10,000m would make you travel at 440m/s in addition to your 300m/s cruise speed, so you need to lose a lot of energy on descent to land at 50m/s. That scales with mass, whereas the speedbraking force doesn't.
It's a little bit more complex with an aircraft, as if it's heavier, it needs more lift, which means a different speed, height or attitude to fly level. More lift usually means more drag, both in approximate proportion to weight, and so more power is required to cruise heavier.
When losing energy descending though, if you try to shed speed deliberately with speedbrakes, that amount of braking isn't affected by weight, just by the speed and spoiler setting, and so a heavier aircraft will need more energy throwing away, and a longer deployment. The energy required to drop from 10,000m would make you travel at 440m/s in addition to your 300m/s cruise speed, so you need to lose a lot of energy on descent to land at 50m/s. That scales with mass, whereas the speedbraking force doesn't.
Join Date: Jul 2009
Location: France - mostly
Age: 84
Posts: 1,682
Likes: 0
Received 0 Likes
on
0 Posts
Originally Posted by awblain
If you load up the same truck with more weight, keeping the same brakes, then it takes longer to stop.
(a) the brake pressure equals the maximum brake pressure, or
(b) the tires start skidding.
In this context 'brakes sufficiently powerful' means that condition (b) applies. The braking force is then proportional to the load on the wheel.
Last edited by HazelNuts39; 22nd Nov 2012 at 21:22.
Join Date: Nov 2004
Location: uk
Posts: 1,224
Likes: 0
Received 0 Likes
on
0 Posts
We're not talking forces or accelerations here, we are talking inertia which is velocity and mass related.
Remember the equation
M1V1=M2V2 (numbers should be in subscript)
Therefore if M1 > M2 then V1 < V2
Remember the equation
M1V1=M2V2 (numbers should be in subscript)
Therefore if M1 > M2 then V1 < V2
Join Date: Aug 2007
Location: Derby UK
Age: 59
Posts: 112
Likes: 0
Received 0 Likes
on
0 Posts
(a) the brake pressure equals the maximum brake pressure, or
(b) the tires start skidding.
(b) the tires start skidding.
The friction of Carbon Carbon brakes increases as they get hotter, so for the same brake pressure as they heat up they get grippier.
(b) not true, Depending on the Mu and the surface texture the maximum retardation is just before the skid onset.
Join Date: Jul 2009
Location: France - mostly
Age: 84
Posts: 1,682
Likes: 0
Received 0 Likes
on
0 Posts
Bye,
How do your remarks make untrue what I wrote, that the maximum braking force is determined by either the maximum that the brake can produce, or the maximum that the tire/runway friction can produce, and that the latter is proportional to the load on the wheel?
How do your remarks make untrue what I wrote, that the maximum braking force is determined by either the maximum that the brake can produce, or the maximum that the tire/runway friction can produce, and that the latter is proportional to the load on the wheel?
Last edited by HazelNuts39; 22nd Nov 2012 at 22:05.
Join Date: Aug 2007
Location: Derby UK
Age: 59
Posts: 112
Likes: 0
Received 0 Likes
on
0 Posts
because that is not what you wrote.
What you did actually write is not true.
My remarks do not make what you wrote untrue, what you wrote that is untrue makes it untrue. I have merely pointed out that what you wrote is untrue.
GB
What you did actually write is not true.
My remarks do not make what you wrote untrue, what you wrote that is untrue makes it untrue. I have merely pointed out that what you wrote is untrue.
GB
Join Date: Dec 2007
Location: Pasadena
Posts: 633
Likes: 0
Received 0 Likes
on
0 Posts
Ms 39,
You're right - I was assuming progressive braking: dumping of heat in the truck brakes at some acceptable level without locking the wheels. If you keep maximum braking force applied, limited by rubber-road friction, then the braking force would rise linearly with weight (until you ripped the tread off the wheel or set fire to the brakes).
You're right - I was assuming progressive braking: dumping of heat in the truck brakes at some acceptable level without locking the wheels. If you keep maximum braking force applied, limited by rubber-road friction, then the braking force would rise linearly with weight (until you ripped the tread off the wheel or set fire to the brakes).