Density altitude question
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tgrayson
New Member
It doesn't cause an increase in TAS, it just requires an increase in TAS in order for your airspeed indicator to register a particular IAS. In other words, your airplane must go faster than 100 knots TAS in order for your airspeed indicator to register 100 knots IAS due to the reduced air density.
tgrayson
New Member
Nope, because the static port pressure isn't affected by density, only pressure. What the pitot tube measures is total pressure, which not only includes the static pressure, but also the pressure caused by slowing the moving airstream to a standstill. That extra pressure (dynamic pressure) depends not only on the air velocity, but also its density.
fish314
Well-Known Member
The following explanation is VERY dumbed down and VERY conceptual (or in other words, pretty sloppy and inaccurate). But it is good for giving you the picture:
Imagine your pitot tube. It is basically just a long tube with a wall at the end called a diaphragm. That diaphragm is attached to a spring, and as the diaphragm is pushed, the spring moves and turns a dial. The dial indicates the airspeed.
What moves the diaphragm in the first place? Molecules of air. As the airplane moves through the air, molecules of air travel down the tube and slam into the wall (diaphragm) at the end. The faster the molecules are moving, the more force they transmit to the diaphragm and the greater the reading on the airspeed indicator.
Of course, other than speed, the NUMBER of molecules hitting the diaphragm also plays a part in how much force is transmitted to the diaphragm. If one air molecule moving at 100 knots transmits X amount of force, then 2 air molecules transmit twice that amount.
Well air density is a measure of the number of molecules in a given volume of air. At sea level, the density is greater than at altitude. Air speed indicators are calibrated for sea level, so at sea level (density altitude) true airspeed and indicated are the same. For sake of argument, let's imagine that to indicate 100 knots at sea level, there are 10,000 air molecules hitting the diaphragm per second, each one moving at 100 knots true.
If we go up in density altitude, there are fewer air molecules in the same amount of volume. Basically the air gets thinner. So imagine we climb to an altitude where there are only 500 air molecules hitting the diaphragm per second, then they would need to be going twice as fast to still get the same amount of force on the diaphragm to register 100 knots. Or, in other words, to indicate 100 knots in this little example, the molecules would have to go 200 knots true (because there are half as many of them hitting the diaphragm).
Does that make sense?
Imagine your pitot tube. It is basically just a long tube with a wall at the end called a diaphragm. That diaphragm is attached to a spring, and as the diaphragm is pushed, the spring moves and turns a dial. The dial indicates the airspeed.
What moves the diaphragm in the first place? Molecules of air. As the airplane moves through the air, molecules of air travel down the tube and slam into the wall (diaphragm) at the end. The faster the molecules are moving, the more force they transmit to the diaphragm and the greater the reading on the airspeed indicator.
Of course, other than speed, the NUMBER of molecules hitting the diaphragm also plays a part in how much force is transmitted to the diaphragm. If one air molecule moving at 100 knots transmits X amount of force, then 2 air molecules transmit twice that amount.
Well air density is a measure of the number of molecules in a given volume of air. At sea level, the density is greater than at altitude. Air speed indicators are calibrated for sea level, so at sea level (density altitude) true airspeed and indicated are the same. For sake of argument, let's imagine that to indicate 100 knots at sea level, there are 10,000 air molecules hitting the diaphragm per second, each one moving at 100 knots true.
If we go up in density altitude, there are fewer air molecules in the same amount of volume. Basically the air gets thinner. So imagine we climb to an altitude where there are only 500 air molecules hitting the diaphragm per second, then they would need to be going twice as fast to still get the same amount of force on the diaphragm to register 100 knots. Or, in other words, to indicate 100 knots in this little example, the molecules would have to go 200 knots true (because there are half as many of them hitting the diaphragm).
Does that make sense?
Fish, Grayson
Thanks, I do understand how the ASI functions. What I'm trying to understand here is: why the decrease in air density doesn't also effect the static port in the same manner that it effects the pitot tube. You say that with a higher DA there are less molecules of air acting on the diaphragm within the pitot tube, correct? Well, then why doesn't the static port also detect a decrease in air density? That's the part I can't wrap my head around. I know the static port measures air pressure. So wouldn't a decrease in air density appear as a lower air pressure to the static port?
In reading these responses, I'm now realizing that no, it would not. So the static port only measures the pressure of the air above it, regardless of air density. But that lower density WOULD be felt by the pitot tube in that it would take a greater amount of ram air pressure to result in a given IAS. Correct?
Thanks, I do understand how the ASI functions. What I'm trying to understand here is: why the decrease in air density doesn't also effect the static port in the same manner that it effects the pitot tube. You say that with a higher DA there are less molecules of air acting on the diaphragm within the pitot tube, correct? Well, then why doesn't the static port also detect a decrease in air density? That's the part I can't wrap my head around. I know the static port measures air pressure. So wouldn't a decrease in air density appear as a lower air pressure to the static port?
In reading these responses, I'm now realizing that no, it would not. So the static port only measures the pressure of the air above it, regardless of air density. But that lower density WOULD be felt by the pitot tube in that it would take a greater amount of ram air pressure to result in a given IAS. Correct?
tgrayson
New Member
Pressure and density are related, but that doesn't matter. The influence of the static pressure is removed from the pitot tube by the airspeed indicator. That's why we have a static port. Your pitot tube measures total pressure, which is the sum of the static and dynamic pressures. The airspeed indicator removes the influence of the static pressure, so its relationship to density is immaterial. The remaining pressure, dynamic pressure, is directly proportional to air density.
fish314
Well-Known Member
hmmmm.....
I don't have a great physical explanation or analogy for this one, so I'm going to have to resort to math. Sorry!
The theory is that total pressure along a stream line is constant (provided you don't add energy to the flow). However, although the total pressure is constant, it comes in two forms: static pressure, and dynamic pressure. Total pressure is static plus dynamic.
Pt=P0+1/2 rho*V^2 (total pressure = static pressure or P0, plus 1/2 the density times the velocity squared).
The airspeed indicator measures total pressure at the pitot tube, and static pressure at the static port. Putting static pressure on one side of the diaphragm and total pressure on the other means that the only movement on the diaphragm is due to the difference between the two.
In other words, it measures Pt-P0. Looking at the formula above Pt-P0=1/2 rho*V^2.
So if we are looking for velocity we either need to factor out the density (to wind up with true airspeed), or we accept that our airspeed contains a density dependence. Well, in a mechanical instrument there really isn't an easy way to get rid of the density term.
Now, there are some electronic airspeed instruments that are capable of correcting for the density term and displaying true airspeed. They are called air data computers, and they are present on most big aircraft. But even still, indicated airspeed is a more useful term for most aviation applications.
Why? Well let's look at the lift equation:
Lift=CL*1/2*rho*V^2*S (lift equals the coefficient of lift times 1/2 times the density times velocity squared times surface area of the wing).
Drag is a similar formula. So you see, lift and drag both depend on the indicated airspeed (V^2 times rho). Therefore things like an aircraft's stall speed are constant if you look at them in terms of indicated airspeed. If you were to talk in terms of true, instead, then at every altitude the aircraft would have a different airspeed.
Same goes for approach speeds, Vne, maximum flap and gear speeds, etc., etc. They are all constant if you look at them in terms of indicated. If you were to look at them in terms of true, they would all vary with altitude.
I don't have a great physical explanation or analogy for this one, so I'm going to have to resort to math. Sorry!
The theory is that total pressure along a stream line is constant (provided you don't add energy to the flow). However, although the total pressure is constant, it comes in two forms: static pressure, and dynamic pressure. Total pressure is static plus dynamic.
Pt=P0+1/2 rho*V^2 (total pressure = static pressure or P0, plus 1/2 the density times the velocity squared).
The airspeed indicator measures total pressure at the pitot tube, and static pressure at the static port. Putting static pressure on one side of the diaphragm and total pressure on the other means that the only movement on the diaphragm is due to the difference between the two.
In other words, it measures Pt-P0. Looking at the formula above Pt-P0=1/2 rho*V^2.
So if we are looking for velocity we either need to factor out the density (to wind up with true airspeed), or we accept that our airspeed contains a density dependence. Well, in a mechanical instrument there really isn't an easy way to get rid of the density term.
Now, there are some electronic airspeed instruments that are capable of correcting for the density term and displaying true airspeed. They are called air data computers, and they are present on most big aircraft. But even still, indicated airspeed is a more useful term for most aviation applications.
Why? Well let's look at the lift equation:
Lift=CL*1/2*rho*V^2*S (lift equals the coefficient of lift times 1/2 times the density times velocity squared times surface area of the wing).
Drag is a similar formula. So you see, lift and drag both depend on the indicated airspeed (V^2 times rho). Therefore things like an aircraft's stall speed are constant if you look at them in terms of indicated airspeed. If you were to talk in terms of true, instead, then at every altitude the aircraft would have a different airspeed.
Same goes for approach speeds, Vne, maximum flap and gear speeds, etc., etc. They are all constant if you look at them in terms of indicated. If you were to look at them in terms of true, they would all vary with altitude.
fish314
Well-Known Member
T, how'd you manage to say in two sentences what it took me like a page and a half to say?
With a higher density altitude, it will require a faster TAS to produce a given IAS.
To remember the effect of DA on the airspeed indicator, you can think of two types of air: thick and thin. Thick air (which is closer to the ground, lower DA) impacts the pitot tube producing a greater force than thin air (which is higher above the ground, higher DA).
Bernoulli Fan
Controller
Yes it does. The static port does "detect" a decrease in air density, but the pitot tube "detects" the same decrease. This means the static pressure ratio between the two is the same no matter the DA. All that's left, then, is dynamic pressure measured by the pitot.
This is just another way of reiterating tgrayson's explanation.
Let's see:
d = density
v = Speed KTAS
Total pressure [P] = Dinamic pressure [q] + Static pressure [p]
q = P - p <<< this takes place on the airspeed indicator. q is shown on the ASI by substracting total air pressure to static pressure
q = 1/2 d v^2 <<< dynamic pressure is subject to these factors
So
1/2 d v^2 = P - p
V = (sqr root of) 2(P-p)/d
A decrease in density will increase the TAS for an object moving at a certain velocity. An object moving at 100KIAS/KTAS at ISA MSL, continues to fly at 100KTAS as it climbs. The decrease in density will make the IAS go down.
Your question actually got me thinking. Maybe it confused me more than it got me thinking lol. I guess static pressure does not depend on air density. Setting the right pressure on your altimeter while sitting on the airport that provided you that setting will show the correct altitude regardless of the air density. An extreemely cold day or a hot day, with the altimeter set to the current pressure will show the field elevation. Now when you start climbing thats another story and the famous temperature/altitude correction chart comes along
d = density
v = Speed KTAS
Total pressure [P] = Dinamic pressure [q] + Static pressure [p]
q = P - p <<< this takes place on the airspeed indicator. q is shown on the ASI by substracting total air pressure to static pressure
q = 1/2 d v^2 <<< dynamic pressure is subject to these factors
So
1/2 d v^2 = P - p
V = (sqr root of) 2(P-p)/d
A decrease in density will increase the TAS for an object moving at a certain velocity. An object moving at 100KIAS/KTAS at ISA MSL, continues to fly at 100KTAS as it climbs. The decrease in density will make the IAS go down.
Your question actually got me thinking. Maybe it confused me more than it got me thinking lol. I guess static pressure does not depend on air density. Setting the right pressure on your altimeter while sitting on the airport that provided you that setting will show the correct altitude regardless of the air density. An extreemely cold day or a hot day, with the altimeter set to the current pressure will show the field elevation. Now when you start climbing thats another story and the famous temperature/altitude correction chart comes along
fish314
Well-Known Member
LV,
The first part is correct, but the two paragraphs beneath the math are not quite right. In your first paragraph, second sentence, you said that, "an object moving at 100KIAS/KTAS at ISA MSL, continues to fly at 100 KTAS as it climbs. The decrease in density makes the IAS go down." This would be true if airplanes held True airspeed constant in the climb, but they don't. Remember that lift and drag are also affected by density in the exact same way that the indicated to true relationship is. Therefore, we hold Indicated constant in the climb, not true airspeed (think small airplanes and lower altitudes, here, because it gets a little more complicated above 10,000 feet and at higher true airspeeds, because some mach effects start to play a role also). Other than that, I think you've got the relationship.
Second paragraph, you mentioned that static pressure is not affected by density, and again that is not true either. There are several different mathematical versions of the ideal gas law, depending on your field of study, but they all say the same thing. They just break the variables up a little differently in chemistry than they do in aeronautical engineering. The way it is usually expressed by aero engineers is:
P=rho*R*T
Or
Pressure (static)= density times the ideal gas constant, R, times Temperature.
As you can see from the equation, pressure just depends on density and temperature (since R is a constant). Or you could say density depends on pressure and temperature, or temperature depends on pressure and density... It's all saying the same thing.
So when you set the pressure correction in the Kohlsman (spelling?) window, you have corrected the instrument for the pressure changes that day. Well from the equation above, that is the same as correcting for temperature and density.
The problem is, both pressure and temperature decrease as altitude decreases. In fact, this is the very principle upon which the altimeter is based. It reads the pressure decrease, and translates it into an altimeter increase. The question becomes, what happens if the rate of decrease on a particular day changes from the rate at which they normally decrease? For example, we know that the standard temperature lapse rate is 2 degrees Celsius (3.6 F) decrease for every 1000' of altitude increase. But it can change. Generally, the pressure lapse rate is pretty steady, because from the equation above, we can see that is only affected by the density lapse rate (very constant) and the by the temperature lapse rate (can vary). In fact, the relationship between pressure and true altitude can also be affected by the actual temperature, as well, as opposed to the temperature lapse rate.
Now, it isn't really a problem most of the time, because most of the time the temperature effects would cause one's indicated altitude MSL to be greater than the actual TRUE altitude. This happens any time the temperature at sea level is above standard day (15C), and it also happens any time the lapse rate is above normal, like in an inversion. Inversions account for almost all of the lapse rate deviations, and warmer than average days make up half of the deviations from standard day (otherwise something ELSE would be the average, right?). In those circumstances, the altimeter reads lower than the actual true altitude. Not really a problem, because that means terrain clearance for everyone is higher than planned. Since all the airplanes are affected in exactly the same way in a particular piece of airspace, that makes traffic separation no problem either.
But on cold days, especially days below freezing, then temperature will cause the altimeter to read higher than the aircraft's actual altitude. Another way to say this is that the airplane will actually be lower than the altimeter reads. Again, all the planes are affected the same way, so de-conflicting traffic is not affected. But terrain clearance is. So we apply a correction factor, especially when we are low to the ground. The rule is you apply the factor to any altitude inside the final approach fix on an instrument approach. Personally, if I were doing a low level flight (say between 0 and 1000' AGL cruise legs), I would also apply the correction there too... although low level I'm going to be mostly using visual and radio altimeter to check my AGL altitude, rather than worry about MSL altitude, but when I cross check my altitude against a chart for backup, I'm going to add in that correction.
Make sense?
Fly_Unity
Well-Known Member
Think of a space shuttle. In space its indicated airspeed is zero because there is no air to register on the pitot tube, however true airspeed (groundspeed) is thousands of miles per hour.
of course in a piston airplane, the TAS wont increase like it does in a rocket because it needs air to work. This is where a turbo charger comes in handy.
of course in a piston airplane, the TAS wont increase like it does in a rocket because it needs air to work. This is where a turbo charger comes in handy.