[aelkins]

I was reading about the different types of terminal velocity.
I ran across the amazing feats of 'Joseph Kittinger'.
It's a great 3 minute read, if you're interested.. http://en.wikipedia.org/wiki/Joe_Kittinger

That made me wonder.. what is the terminal velocity of an average RV?
Assuming 1100 lbs Gross, free-spinning FP prop, sufficient altitude available.
So, if you pulled the throttle to idle, pushed the nose straight down.. at what approximate speed would the drag equal gravity on a slick RV?

Just curious..

 

[FlyerJumper]

Likely never know...

Best guess...
The wings would come off first!

 

[RScott]

Then terminal velocity might be even higher! At that point, I don't think I would care what the TV was.

 

[jonbakerok]

No idea, but about Col. Kittinger

I took my folks up to Bartlesville for the Biplane fly-in once and there was guy giving rides in a "New Standard" biplane. I bought a rides for my whole crew. We had a great ride and my dad took pictures of us all in the cockpit.

Years later, I was reading about Col. Kittinger's exploits in some magazine and the story mentioned that in his retirement he did all kinds of things to stay active in aviation -- such as spending a summer barnstorming in a "New Standard Biplane". I thought, no, it couldn't be. But I got out the pictures and sure enough, there he was, with "Col. Kittinger" written right on the rear cockpit. Man, I got to fly with one of the great's and didn't even know it.

 

[tonyjohnson]

Col Joe Kittinger is still flying

Col Kittinger lives in Orlando, the town where he grew up. For the past 25 years I have been proud to call him a friend. He is a great guy, a patriotic American, an man of integrity and honor, and of course a superb aviator.

Not only does he hold the record for the highest parachute jump, 102,800 feet, was the first and I think only, man to break the sound barrier without an airplane (during the jump), he is also the first to fly a balloon solo across the Atlantic. He did 4 tours in Vietnam, 3 as a pilot and the 4th as a POW in the Hanoi Hilton after his F4 was shot down. He was the senior prisoner, as a Colonel. He caused his captors so much trouble they transfered him to another area.

He is not barnstorming in the New Standard now, but still flies. He and I flew my Taylorcraft L2 about a month ago.

He and his wife Sherry, who is a lovely and sweet person, will be at Sun and Fun.

Col Kittinger is one heck of a guy. The world needs more like him.

Tony

 

[ScottSchmidt]

Terminal Velocity

Does anyone know the coefficient of drag for an RV?
Once we have that number and I have a number for the surface area from an aircraft design book for an RV-4. We then need to plug that into the equation:
mg=1/2(Fl)(v^2)(S)(cd)

 

[keen9a]

Some of the CAFE reports have Cd. The reports are here:
http://cafefoundation.org/v2/research_aprs.php

 

[strahler13]

Ben, the Chief is gone. Illinois is a great school, but unfortunately it, like most schools of higher lernin' are governed by a bunch of "weenies". No backbone. More interested in being "PC" than caring about the wishes of the majority of their alumni. Sorry, just a little OT.
Mark

 

[rv6ejguy]

Dive tests with no prop were performed on a P51, towed aloft by a Black Widow. Reached something close to mach .75 apparently!

 

[gmcjetpilot]

Does anyone know the coefficient of drag for an RV?
Once we have that number and I have a number for the surface area from an aircraft design book for an RV-4. We then need to plug that into the equation:
mg=1/2(Fl)(v^2)(S)(cd)

where

Vt is the terminal velocity,
m is the mass of the falling object,
g is gravitational acceleration at the Earth's surface,
Cd is the drag coefficient,
ρ is the density of the fluid the object is falling through, and
A is the object's cross-sectional area.


m = 1,600 lb mass
g = 32.174 feet/sec^2
Cd =0.021 (zero lift)
A = 2.32 ft^2 (RV-6A)
ρ = 2.3768908 x 10^(−3) (slugs/Ft^3)

Plug and Chug and turn the crank

So V = ((2*m*g)/(ρ*A*Cd))^0.5 = 29,817 mph


One factor not included, besides Vne and wings coming off, the ground. The ground will come up and "Smite Ye Mightily".

 

[Mark Burns]

That's pretty fast

Plug and Chug and turn the crank

So V = ((2*m*g)/(ρ*A*Cd))^0.5 = 29,817 mph


One factor not included, besides Vne and wings coming off, the ground. The ground will come up and "Smite Ye Mightily".[/QUOTE]

29,817 MPH !!!
I'm not an engineer but I'm sure it would never go that fast. I suspect it would burn up like a meteorite if it did.

After the wings came off I think the fuselage would break up at about 500 and then all but the engine would fall much more slowly. I don't see the engine by itself going over 1500 :eek:

I may regret posting this. :D Anyway, that's what I think.

Mark

 

[airguy]

One factor not included, besides Vne and wings coming off, the ground. The ground will come up and "Smite Ye Mightily".

Not if you pull 837 gee's and pull out of the dive....

 

[gmcjetpilot]

Yep, garbage

29,817 MPH !!!
I'm not an engineer but I'm sure it would never go that fast. I suspect it would burn up like a meterorite if it did. Mark

You think? Yep Mark I don't think that's correct. Not sure what limits terminal velocity would be say off the airframe was infinitely strong and you started say at 90,000 feet. In engineering some solutions are not valid. Might have made a math error?

Still one thing we can learn is pointing a RV nose down will cause the airspeed to accelerate well beyond Vne and structural failure or flutter will occur within a hand full of seconds.

I heard of a scary RV story, where a guy let his buddy fly his RV. From cruise he rolled upside down and went into a split-S (dive). Before he pulled out (at about +9 g's) the elevator fluttered. They survived, but there where wrinkles in the wing and the stabilizer and elevators where damaged. Lesson, a split-S started faster than 110-130 mph and low/no power may get you hurt.

However the formula you can see mass is key; thats why you need to be careful near the ground. A large jet 250,000 lbs or 400,000 lbs with low drag needs extra attention near the ground. A recovery from any "unusual attitude" may take 10,000 feet or more.

 

[N941WR]

Cross section?

George,

Is that X-section number correct? Does the RV really have that little of a cross section, including the wings?

 

[gmcjetpilot]

Did not check

George,

Is that X-section number correct? Does the RV really have that little of a cross section, including the wings?

Oh yea they left the gear out I think? I just copied it, I did not check it. There is probably a mistake. I wounder if I needed to divide mass by 32 ft/s^2 for "slugs" (really that's the term). Even if I played with mass, slugs and flat plate area I would get 4,000 to 5,000 mph, also a ridiculous number since the wings or tail would be gone by 300 mph?

 

[szicree]

where

Vt is the terminal velocity,
m is the mass of the falling object,
g is gravitational acceleration at the Earth's surface,
Cd is the drag coefficient,
ρ is the density of the fluid the object is falling through, and
A is the object's cross-sectional area.


m = 1,600 lb mass
g = 32.174 feet/sec^2
Cd =0.021 (zero lift)
A = 2.32 ft^2 (RV-6A)
ρ = 2.3768908 x 10^(−3) (slugs/Ft^3)

1600 lbs is not mass, it's weight. The corresponding mass would actually be 1600/32.174 = 49.72 slugs. Or, more simply, just use 1600 for the mg. Oh, and the formula is in feet/sec, so you'll need to convert to mph. It appears that the correct value is 3584.14 mph. I still ain't buyin' it though. There are very few things that'll move through air that fast under gravity alone.

 

[airguy]

This formula does not take compressibility and mach drag into effect - it's strictly for subsonic values.

 

[Pirkka]

I prefer SI-units to avoid problems so this is what I got:

Vt = ?
m = 1600 lbs * 0.453 592 37 kg/lbs= 726 kg
g = 9.81 m/s
Cd = 0.021
p = 1.2 kg/m^3
A = 0.216 m^2

Units:
sqrt((kg * m / s^2) / ((kg / m^3) * (m^2))) = sqrt(m^2 / s^2) = m/s

So units does match... then the numbers would yield: 1144 m/s which equals 4118 km/h which would be 2559 mph... Would this be closer the truth? Pretty fast anyway.

 

[Kevin Horton]

Cd is the drag coefficient,
A is the object's cross-sectional area.

Cd =0.021 (zero lift)
A = 2.32 ft^2 (RV-6A)

That looks like the zero-lift CD reported by the CAFE Foundation for the RV-6A. It is referenced to the wing area, not the flat plate area. So, use 110 ft^2, rather than 2.32 ft^2. And, if the speed is greater than about mach 0.7, the CD will increase very quickly, so the solution becomes invalid.

 

[rzbill]

Mach Sprach

This formula does not take compressibility and mach drag into effect - it's strictly for subsonic values.

I KNEW there was a reason to buy that ancient Mach meter at the Fly Mart! I wonder how it will fit next to the glass panel du jour?

 

[Auburntsts]

I agree with Kevin. I think the frontal area is off, and I also believe the Cd is too low as well. A couple of 1/100ths can make a 300 ft/sec difference. My calc shows the Vt to be around 430MPH given a frontal area of 110 sq/ft and a Cd of.03 (still way low in my opinion for the whole airframe). I think the only way you'd see supersonic velocities would be in a vacuum (or in a near vacuum like the upper stratosphere) or if you tripled the mass.

Todd
#40631
RV-10 Emp
Dodging mortars in Iraq instead of building

 

[burgundyja]

i am just going to guess and say 280mph. with the engine off but still windmilling. i heard somewhere that on a c-172 it is 150kts

 

[Auburntsts]

I think the 150kts on a C-172 is Vne (148KTS maybe) which has no relation to Vt, except if you exceed Vne you may soon learn what Vt is for your configuration.

Todd

 

[szicree]

I prefer SI-units to avoid problems so this is what I got:

Vt = ?
m = 1600 lbs * 0.453 592 37 kg/lbs= 726 kg
g = 9.81 m/s
Cd = 0.021
p = 1.2 kg/m^3
A = 0.216 m^2

p cannot be correct. This would mean that a cubic meter of air would weigh about 2.6 pounds! Also, g should be in m/s^2,

 

[Mike S]

My estimate

Warp factor .000001??

Mike

 

[longranger]

p cannot be correct. This would mean that a cubic meter of air would weigh about 2.6 pounds! Also, g should be in m/s^2,

At standard temperature and pressure (sea level, 15C (59F) a cubic meter (35.3 cu ft) of air weighs closer to 2.7 pounds.

Airguy and Kevin Horton hit the nail on the head; the forumula given is only valid for relatively low subsonic velocities.

Miles

 

[Pirkka]

p cannot be correct. This would mean that a cubic meter of air would weigh about 2.6 pounds! Also, g should be in m/s^2,

We are looking for a density of air right? http://en.wikipedia.org/wiki/Density_of_air And the g is m/s^2 when I was calculating the units but not upper as it was post to be.

For others:

I agree that 2.32 ft^2 is way too small for whole plane but also 110 ft^2 is way too much if we are talking about nose towards ground situation. That is like 10 x 11 ft, how would you fit that area as sectional area of your plane? I quess 10 - 20 ft^2 would be much much closer to the truth.

Somebody also questioned Cd. Wikipedia gives some Cd values for for planes http://en.wikipedia.org/wiki/Drag_coefficient

• 0.027 - Cessna 172/182
• 0.027 - Cessna 310
• 0.022 - Learjet 24
• 0.048 - F-104 Starfighter
• 0.021 - F-4 Phantom II (subsonic)
• 0.044 - F-4 Phantom II (supersonic)
• 0.031 - Boeing 747
• 0.095 - X-15

Edit: now I found out where that 100 ft^2 comes from. Cafe report as RV-6 wing area... but according to the same report, the flat plate area should be that 2.32 ft^2. Pretty low... but the frontal sectional area or RV ain't as the "outlines" as when air is coming from front the most shapes are round and therefore if aerodynamic shape is converted to the flat plate it will be "smaller".

Ok... to solve this issue we need a brave test pilot and plane... RV...

 

[mgomez]

It's about 550 MPH

If you ignore the drag of windmilling prop (which is hard to calculate), and use the 2.3 ft2 equivalent flat plate area for an RV-6 or -7, you get a terminal velocity of ~550 mph.

To verify this speed, try this math:

- Top speed of a 200 HP RV-7 = 216 MPH
- 1 HP = 550 ft-lbs/s
- Power = drag * speed, so drag = Power/speed
- Assume the prop is 80% efficient
- 216 MPH = 317 ft/s
- drag @ 216 MPH = 278 lbs

Now assume that the airplane is going vertically, so that drag = weight.
terminal speed = sqrt(1800/278) * 216 = 550

And the density of air under the standard atmosphere at sea level IS 1.23 kg/m3, which is about 2.7 lbs/m3. For those of you who prefer oddball units, that's 0.002378 slug/ft3.

 

[szicree]

We are looking for a density of air right? http://en.wikipedia.org/wiki/Density_of_air And the g is m/s^2 when I was calculating the units but not upper as it was post to be.

Sorry, I stand corrected; although I still can't get my head around the idea that there are 2.6 lbs of air in a 3 foot by 3 foot box. Heck, this means that the air in the box weighs more than the box?! I'll accept the number as correct, but I just don't get it.

 

[Ironflight]

Old Geezer Math...

OK guys, I am enjoying the heck out of watching all of these calculations, because I know that many of you (us) are engineers...many are aeronautical engineers....and we are still all coming up with different numbers! It just goes to show that school doesn't stick with you very long when you get out in the real world! I specialized in aircraft design and flight testing way back when I was a student, but I can't remember the last time I did any basic stuff like this....

So y'all might as well laugh at me as well, here's my take (rememebr, I have young people that do this for me these days - with computers!):

The classic drag equation of "one half rho vee squared times Cd times area" is (or at least was, when I was a student) based on wing area. If I use 100 square feet for an RV, the good old rho of .002378 (one of the few numbers I remember other than my SSN), and convert from fps to knots, I get an equation for V = 0.59 * ((13456/Cd)^0.5). Plugging that into a spreadsheet and solving for a range of Cd's from 0.01 to 0.04 gives me terminal velocities ranging from 685 knots to 340 knots, respectively. And yup, compressibility is going to be significant above 0.5 mach, and dominate much above that.

For whoever we decide gets to go try it (I vote for George... ), they are going to have to point it straight down and then control pitch very carefully to maintain zero lift so as not to add drag and become the proverbial chicken in a hurricane, with it's feathers blowing off!

All in good fun.....(and my pure guess is that you'll max out about 300 - 350 knots).

Old, worn out aero engineer....and now you've cost me my lunch time and given me a headache!

Paul

 

[osxuser]

The 550mph number sounds reasonable (relitively). I've heard of a guy going vertical with the engine off, CS prop windmilling at 2700 and the airplane staying below VNE. Whether this is true is hard to say, but it's possible with something draggy like a 182. RV would probably not work quite the same.

 

[rv6ejguy]

If you ignore the drag of windmilling prop (which is hard to calculate), and use the 2.3 ft2 equivalent flat plate area for an RV-6 or -7, you get a terminal velocity of ~550 mph.

To verify this speed, try this math:

- Top speed of a 200 HP RV-7 = 216 MPH
- 1 HP = 550 ft-lbs/s
- Power = drag * speed, so drag = Power/speed
- Assume the prop is 80% efficient
- 216 MPH = 317 ft/s
- drag @ 216 MPH = 278 lbs

Now assume that the airplane is going vertically, so that drag = weight.
terminal speed = sqrt(1800/278) * 216 = 550

And the density of air under the standard atmosphere at sea level IS 1.23 kg/m3, which is about 2.7 lbs/m3. For those of you who prefer oddball units, that's 0.002378 slug/ft3.

The prop will add a lot of drag at this speed and other bad things happen in practice. Some of the terminal dive tests on the Spitfire in 1945-1946 had the prop and reduction gear leave the airframe due to severe over speeding and reverse loading. The P51 tests were designed to quantify propeller drag so tests were done with and without prop. The pilot reported an eerie experience as far as the sound went, just air whistling over the airframe and the whir of recording cameras. Mach .75 is pretty impressive with no engine. The Shuttle of course has a head start with the altitude it starts at.

 

[RVbySDI]

Gotta love engineers

I really love all of you engineers and mathematicians who can remember all of these various formulas. The idea of a discussion about an event that, if was to occur in real life, would cause my demise were I unfortunate enough to find myself in such a predicament is very entertaining reading.

I am not quite sure where the benefit is in knowing what this terminal velocity number is but please keep posting as this discussion is very fascinating.

 

[Ironflight]

The Shuttle of course has a head start with the altitude it starts at.

Yeah...the Shuttle doesn't accelerate to terminal velocity, it decelerates to it!

 

[frankh]

All I know is

MY body's terminal velocity is 120MPH lying flat and about 200mph in a "stand-up" or head down vertically!

Ha...I can almost keep up with my RV straight down...

Frank...ex skydiver and 7a driver

 

[AlexPeterson]

If you ignore the drag of windmilling prop (which is hard to calculate), and use the 2.3 ft2 equivalent flat plate area for an RV-6 or -7, you get a terminal velocity of ~550 mph.

To verify this speed, try this math:

- Top speed of a 200 HP RV-7 = 216 MPH
- 1 HP = 550 ft-lbs/s
- Power = drag * speed, so drag = Power/speed
- Assume the prop is 80% efficient
- 216 MPH = 317 ft/s
- drag @ 216 MPH = 278 lbs

Now assume that the airplane is going vertically, so that drag = weight.
terminal speed = sqrt(1800/278) * 216 = 550

And the density of air under the standard atmosphere at sea level IS 1.23 kg/m3, which is about 2.7 lbs/m3. For those of you who prefer oddball units, that's 0.002378 slug/ft3.

My calculations agree completely, about 550 mph. For those that didn't follow Martin's calculations, he simply calculated what the constant terms "one half times density times area times coefficient of drag" were, by stating that a 200 hp airplane gets 216 mph (at 200 hp, 80% prop efficiency), and then used this constant to calculate the new velocity. Very elegant way to do it.

It passes the smell test also - my dad told me that the F86, when taken to 45,000', and pointed straight down, would break the speed of sound briefly before entering dense enough air to slow back below it. Interestingly, it didn't matter whether the engine was at idle or full power. This is because the horsepower associated with giving up potential energy at those speeds is overwhelmingly larger than the engine's output power. (BTW, many believe that an F-86 broke the sound barrier successfully before the X1, but money/politics prevented it from being announced. Whole different story.)

The 1800 pound RV in a straight down dive at our calculated terminal velocity (obviously assuming the airframe would tolerate it) would be dissipating 2641 hp. Wouldn't slow down much by pulling the throttle to idle.

 

[kentb]

Frank is ready to race George...

MY body's terminal velocity is 120MPH lying flat and about 200mph in a "stand-up" or head down vertically!

Ha...I can almost keep up with my RV straight down...

Frank...ex skydiver and 7a driver

But Frank I think you will start to flutter before catching up.

Kent

 

[rv6ejguy]

My calculations agree completely, about 550 mph. For those that didn't follow Martin's calculations, he simply calculated what the constant terms "one half times density times area times coefficient of drag" were, by stating that a 200 hp airplane gets 216 mph (at 200 hp, 80% prop efficiency), and then used this constant to calculate the new velocity. Very elegant way to do it.

It passes the smell test also - my dad told me that the F86, when taken to 45,000', and pointed straight down, would break the speed of sound briefly before entering dense enough air to slow back below it. Interestingly, it didn't matter whether the engine was at idle or full power. This is because the horsepower associated with giving up potential energy at those speeds is overwhelmingly larger than the engine's output power. (BTW, many believe that an F-86 broke the sound barrier successfully before the X1, but money/politics prevented it from being announced. Whole different story.)

My dad flew the Canadair Sabre with the more powerful Orenda engine. Easily supersonic in a steep dive above 25,000 feet, but if you didn't break mach 1 by then, you never would. Chasing the mach and increasing air density made it impossible down low. I think he had one up to 1.06 in a dive from about 49,000 feet. The slatted Orenda powered F86 was the king of the heap until the F100 showed up. The J47 USAF ones did not come out ahead too often on encounters with Canadian Sabres.

Jet aircraft are pretty different however from prop driven aircraft. Engine overspeeds were pretty common on WW2 fighters in vertical dives from very high altitudes. When you hit the coarse pitch stop at idle throttle, rpm is only gonna do one thing!

 

[mgomez]

It is pretty hard to swallow!

Sorry, I stand corrected; although I still can't get my head around the idea that there are 2.6 lbs of air in a 3 foot by 3 foot box. Heck, this means that the air in the box weighs more than the box?! I'll accept the number as correct, but I just don't get it.

Yeah, it's pretty hard to believe that air is so dense, I'll give you that.

But then again, that same 1 cubic metre box, if filled with water, would weigh 1000 kg, or 2200 lbs...as much as a small airplane. That's pretty hard to believe, too, for me anyway.

 

[kentb]

Something is wrong here...

Yeah, it's pretty hard to believe that air is so dense, I'll give you that.

But then again, that same 1 cubic metre box, if filled with water, would weigh 1000 kg, or 2200 lbs...as much as a small airplane. That's pretty hard to believe, too, for me anyway.

A cubic meter of water ~27 cubic ft would weigh less then 300 lbs.

Kent

 

[Mike S]

Nope.

Apx 7.48 gal/cuft, times 8.6 lbs/gal, times 27 cuft=1736+ lbs. And a metre is more than a yard, so 2000 lbs isnt far off.

Finally get to use old fireman knowledge with all the engineers out there.

Mike

 

[hngrflyr]

My calculator says a cubic meter of water weighs ~2200 lbs. 35.32 cubic feet X 7.5 gallons per cubic foot X 8.3 lbs per gallon = 2198.67 lbs

 

[mgomez]

Now it's really time to go work on my RV

http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html

 

[gmcjetpilot]

All in fun

Its just for fun, that is why engineering is more than formulas and constants and units. I knew the answer was ridiculous and invalid, but got your attention. Besides seriously terminal velocity in a RV is ridiculous, the wings, tail or control surface will come off well before 550 mph. Also work out those speeds, 550 mph straight down = 550 * 88 = 48,400 feet per min. Say you start high up, from 10,000 feet to zero would be 12.5 seconds.

PS: 110 sq-ft is wing area not frontal area.

 

[David-aviator]

The 550mph number sounds reasonable (relitively). I've heard of a guy going vertical with the engine off, CS prop windmilling at 2700 and the airplane staying below VNE. Whether this is true is hard to say, but it's possible with something draggy like a 182. RV would probably not work quite the same.

OK, you engineer guys, here's some terminal velcocity information from a total non engineer based on personal experience. My guess is Paul is in the ball park with 300-350 knots in most any airplane.

A long time ago when I was young (21) and dangerous, and in training in the F-86, we had a lesson to take the airplane through mach one to qualify for the NAA mach buster pin. The lesson plan was to climb to something like 40,000 feet, and with full throttle, push the nose over into an near vertical dive and watch the mach meter. It took some doing. From flight test data in the FHB ( thanks to friend, I have a copy), the airplane could go to Mach 1.17 after a sustained maximum power 90 degree dive. At the time we had about 300 hours total time so they did not want us doing such a demonstration, just mach one, which we of course were glad to do. Also, from the FHB, one plot at 90 degrees and mach .8 started at 240 KIAS at 40,000' and was 380 KIAS at 20,000. This obviously was with some power on. Without power, it would not be 380 knots at 20,000'.

On the other hand, terminal velocity of the Cessna is much higher than 150 knots or whatever was stated. Again, from personal experience - while qualifying for a CFI ticket, a very young CFI took this young 2nd LT. up for the required FAA spin demo. I asked him if he was going to demonstrate the recovery, since we had not even talked about it, and he said no, you have more experience than I, just do it. (little did he know)

For lack of much knowledge of the c-150, I decided to use the T-28 recovery technique. After 3 turns, the stick was popped forward and WOW, did that little C-150 wind up. The airspeed indicator not only went through the red line very quick, it nearly pegged at the end of the numbers. The gentle pull out felt like the wings would come off any moment. (I asked the CFI if he wanted to see another one and he said no that would do. For sure he did not want to risk doing that again!)

I wish I knew more math. Isn't terminal velociy what some guy figured out in the Middle Ages by dropping a rock from a tall tower and timing its descent from various elevations? At some point drag would equal weight and that would be as fast as it is coming down - terminal velocity.

 

[rv6ejguy]

In truth, not likely an RV would actually go 550mph in a dive. Compressibility effects on the thick wing, landing gear and blunt shape would probably stop you at mach .6 or so. I think the Cozy actually has a mach limit around this figure with a thinner wing. The wing section % has a big effect on mach limitations. Spitfire was over .9 because of the thin wing section. P38, P51, Typhoon were all considerably lower with thicker wing sections.

 

[gmcjetpilot]

You too can get a Nobel Prize in Physics

Isn't terminal velocity what some guy figured out in the Middle Ages by dropping a rock from a tall tower and timing its descent from various elevations? At some point drag would equal weight and that would be as fast as it is coming down - terminal velocity.

I think you might be talking about the universal law of gravity, it is constant. Galileo alleged dropped things, but its my favorite physicist and mathematician, Sir Isaac Newton, regarded by many as the greatest figure in the history of science determined the universal laws of gravity.

Galileo's reported experiment from the Leaning Tower of Pisa is arguably the best known of all physics stories. Unfortunately no one quite knows whether it really happened. Evidence that someone did it in Galileo's time. Text indicated test and theory of using two cannon balls, one weighing one or two hundred pounds, or even more, will not reach the ground by as much as a span ahead of a musket ball weighing only half a pound, provided both are dropped from a height of 200 cubits. The idea was to minimize air drag or terminal velocity. Of course this would not work if you where at 10,000 feet, but they where not flying around back than.

The principle is acceleration for all objects are constant. In a vacuum this is true, but for limited heights and similar shapes you could come to this conclusion, a be it a crude experiment. Of course drag from air was a factor, but their test did show acceleration from gravity is constant.

I don't think that Newton discovered terminal velocity, but he did formulated more sophisticated gravity experiments that proved gravity is a constant, the foundation of the universal law of gravity and related to Newtonian Laws of Motion (Newton's First Law, Inertia - objects in motion remain in motion unless acted upon by other forces; Second Law - Force is equal to mass times acceleration; Third Law - every action there is an equal and opposite reaction.)

 

[tonyjohnson]

terminal velocity

You guys have way too much time on your hands.

 

[szicree]

Yeah, it's pretty hard to believe that air is so dense, I'll give you that.

But then again, that same 1 cubic metre box, if filled with water, would weigh 1000 kg, or 2200 lbs...as much as a small airplane. That's pretty hard to believe, too, for me anyway.

I just couldn't get my head around the idea that a 3-foot square box full of air would weigh over two pounds, not counting the weight of the box. But then I remembered that the box of air is submerged in a "sea" of air, and therefore is bouyed up by a force equal to the weight of the air it displaces. In other words, it weighs nothing, in spite of the fact that it has plenty of mass. I feel better now.

 

[mgomez]

I just couldn't get my head around the idea that a 3-foot square box full of air would weigh over two pounds, not counting the weight of the box. But then I remembered that the box of air is submerged in a "sea" of air, and therefore is bouyed up by a force equal to the weight of the air it displaces. In other words, it weighs nothing, in spite of the fact that it has plenty of mass. I feel better now.

Another way to get a feel for the weight of air is this:

The pressure at sea level is 14. 7 psi. That means that every square inch of the ground is holding up a tall column of air that weighs almost 15 lbs. That column extends up to space, obviously. The density is constantly decreasing with height , but half of the weight is due to the lower 18,000'.

So a vertical column of air that's 1" square and 18,000' tall weighs over 7 lbs!

And those who said that the terminal velocity of an RV is less than 550 mph are right. The transonic drag rise and drag of the prop would certainly be significant. I was just trying to point out that it's not tens of thousands of mph.