High Speed Climb
 

Why is it then some airplanes, after t/off and when given high speed
climb reply by saying " ok on passing 5000ft high speed"
Any reason for this? :ok:

Company speed limit to 250 below 5,000 ft

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

Yes you can, up to a point. The retarding force that can be derived from the friction between tires and road increases proportionally to the load that the weight exerts on the road. Provided the brakes of car and lorry are sufficiently powerful and equally efficient, car and lorry will slow down at the same rate, even though their masses are different. If the lorry has a higher aerodynamic drag than the car, it may even slow down faster. In practice that is unlikely, however, because the tires and braking systems of modern cars are more effective/efficient than those of lorries.

But your original question was about aircraft, not cars and lorries, and the physics are somewhat different.

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'm wondering if your simple analogy explains that the heavier aircraft slows down more rapidly than the lighter aircraft below the minimum drag speed.

Regards,
HN

thats aerodynamics as opposed to basic Physics,so no it doesn't and wasn't really intended to, so failing that have a look here.

http://home.anadolu.edu.tr/~mcavcar/htk224/MinDrag.pdf

GB

Bye,

Post #2 states the 'basic physics'.
F/M is 'basic' aerodynamics.
Your analogy doesn't address either one.

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 ?:bored: ?:rolleyes: 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 !);)

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

SMOC states Newton's second law as: F=MA. It relates force F, mass M and acceleration A. Energy doesn't come into it.

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.

It's essentially explained in post #17:
What he writes for speeds above green dot, is reversed below green dot.

Regards,
HN

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.

Not necessarily, it depends on the brakes and the friction available between tires and road surface. That's why I added the caveat 'brakes sufficiently powerful'. The maximum retarding force from a braked, rolling wheel is obtained when one of the following conditions is reached, whichever comes first when increasing the pressure applied to the brakes:
(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.

i think Mr Einstein might be turning in his grave. :\

i give up :ugh::ugh:

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

(a) not true, under extreme hard braking there are small particles of metal embedded into the compound which get hot and "weld" themselves to the brake disc, this forms a mechanical bond which has to be physically broken rather than just relying on friction.

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.

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?

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

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).

If you have, I don't get it. Must be my fault.