Does wind speed and gusts affect descent rate?

[sky12345 0]

Quagmirian

***I vagely remember the episode of mythbusters where they proved that a bullet shot horizontally falls slower than just dropped

Ahh, the problem here is that a canopy's airpseed stays the same regardless of windspeed. The bullet analogy is more suited to a skydiver exiting a moving aeroplane compared to a balloon or a helicopter. In this case, the airspeeds are different, so descent rate is affected.

true
i was only answering format's thought experiment

otoh... by analogy with this experiment, even a round parachute (with zero forward drive) will change its decent rate when going through shear layers, and it will float down always slightly slower regardless of whether u go from faster layer to slower or vice versa

[pchapman 261]

sky12345

i vagely remember the episode of mythbusters where they proved that a bullet shot horizontally falls slower than just dropped

Non-relevant. Bullet is moving relative to the air mass. Aerodynamic forces on the bullet have to be calculated as a whole, and can't purely be calculated purely on horizontal and vertical movement separately without reference to the other. The simple way to put it, is that there are significant lift and drag forces on a high speed object (relative to the air mass, not the earth or solar system) which are different from one dropped from a standing start, and thus affect the time-of-fall.

[kallend 1,619]

sky12345

***If I am to make a maple seed out of aluminium or other material and/or size to match density to water in proportion, as it does have proportion in air...

and it emulates behavior in water exactly like real one in the air...

You are positive that it will hit bottom of, say 2 feet deep pool at the same time as in another pool, same depth, with artificial steady water current (of any speed)?

Give me a "yes" and I quit and thank you for your patience.

the answer is "no"

i vagely remember the episode of mythbusters where they proved that a bullet shot horizontally falls slower than just dropped

Not the same thing at all. And easily proved using regular Newtonian physics.

...

The only sure way to survive a canopy collision is not to have one.

[kallend 1,619]

pchapman

***
i vagely remember the episode of mythbusters where they proved that a bullet shot horizontally falls slower than just dropped

Non-relevant. Bullet is moving relative to the air mass. Aerodynamic forces on the bullet have to be calculated as a whole, and can't purely be calculated purely on horizontal and vertical movement separately without reference to the other. The simple way to put it, is that there are significant lift and drag forces on a high speed object (relative to the air mass, not the earth or solar system) which are different from one dropped from a standing start, and thus affect the time-of-fall.

Because drag goes as v^2 at the Reynolds numbers involved, the problem becomes non linear and the horizontal and vertical components of drag are no longer independent. Easily proved, and I think I posted the proof some years ago in the "freefall drift" discussion on here.

...

The only sure way to survive a canopy collision is not to have one.

[kallend 1,619]

sky12345

******I vagely remember the episode of mythbusters where they proved that a bullet shot horizontally falls slower than just dropped

Ahh, the problem here is that a canopy's airpseed stays the same regardless of windspeed. The bullet analogy is more suited to a skydiver exiting a moving aeroplane compared to a balloon or a helicopter. In this case, the airspeeds are different, so descent rate is affected.

true
i was only answering format's thought experiment

otoh... by analogy with this experiment, even a round parachute (with zero forward drive) will change its decent rate when going through shear layers, and it will float down always slightly slower regardless of whether u go from faster layer to slower or vice versa

But format's thought experiment explicitly excluded shears.

...

The only sure way to survive a canopy collision is not to have one.

[kallend 1,619]

JeffCa

***
And these effects have been exhaustively discussed a decade ago. Quit being lazy, look it up and stop wasting everyone's time.

From one (very recently retired) teacher to another, I question your professional attitude as a physics professor if you have these kinds of comments towards people trying to think out the concepts for themselves, instead of just being lazy and going to the back of the book for the answer. Yes, looking up the answer is the lazy solution here, contrary to what you might think. Seriously, we're trying to consider it for ourselves, there's no harm in it, and you keep coming here telling us to stop and check an old thread. I would never tell my students to stop thinking for themselves and just look it up. They can look it up after they've thought about it themselves. I am aware of the old thread, but have liked the way this new thread forced me to think. If you think we're wasting your time, don't read it.

"No class, we won't be doing the tickertape cart experiment this year, we'll just be looking up what should happen. It would be wasting our time to do or discuss anything that has already been done. Read the lab report from last year's students. No, no, don't try to think about what might happen if a wave passes through a double-slit, that would be lazy and it has already been done. Look it up."

1. If he really wanted to work it out for himself instead of just arguing about it, he could have done that. However, he simply argues with everyone who tells him what's what.

2. He is NOT my student. My students are expected to show initiative in finding stuff for themselves.

3. I responded to the question posed in the thread title by telling the questioner where to find the answer.

...

The only sure way to survive a canopy collision is not to have one.

[kallend 1,619]

sky12345

>Since you can't make a 2' deep pool with steady current of a viscous fluid that has the same speed all the way to the bottom

sure u can
just move the entire pool (with still water in it)
boom! for the object dropped from above in it its tye perfectly steady current of viscous fluid
or just drop the object in nonmoving still pool with same horizontal speed
its relativity
e = mc2

maybe u should take physics 101 mr "physics professor"

Yep. Unfortunately it doesn't model the atmosphere.

...

The only sure way to survive a canopy collision is not to have one.

[sky12345 0]

>Not the same thing at all. And easily proved using regular Newtonian physics.

how a bullet fired horizontally and a bullet dropped in the moving stream not the same thing? the result is the same.. it moves vertically slightly slower than bullet dropped in still medium

>horizontal and vertical components of drag are no longer independent

what do u mean independent? they r just projections of a vector, how they can be independent?

>But format's thought experiment explicitly excluded shears.

can u read b4 replying? when i finished with formats experiment, i introduced my own one, with shears.... reading comp probs?

>Yep. Unfortunately it doesn't model the atmosphere.

again can u read? i was talking about producing uniform stream in formats exp. it has nothing to do with the atmosphere, its a thought experiment!!

i had F- in physics but even an idiot like me knows it better than "physics professor"

hereby i demote u of your title and advise u to go to a middle school and start it from the ground up

[sjc 0]

k_marr08

I know when facing into the wind, ground speed decreases when wind speed increases, but does descent rate decrease as well? Does the added wind speed provide more lift?

In a steady state flight, when nothing changes, there should be no affect of wind speed on descent rate. This is only true if nothing changes at all. At the same time, air density and temperature changes with altitude and these will act as changing inputs.

It helps to think about the parachute and skydiver as two bodies. These bodies have different momentum (different mass) and different drag. When both bodies get the same additional input, their relative speed changes differently and this causes the canopy to surge forward or the skydiver to swing forward. This is similar to what is happening when we apply inputs to the canopy. These changes will affect angle of attack, which will affect lift coefficient and therefore the descent rate. I understand that this formula is for a fixed wing aircraft, but the principle still applies.

k_marr08

And, a related question, if one were to downwind a landing, would their descent rate increase, or would only their ground speed increase?

See above.

k_marr08

If the answer is that the descent rate is NOT effected by wind speed, then why do gusts "feel" like they affect descent rate? Is there some additional element of gusts that actually do affect descent rate?

Gusts affect landings because they cause change in angle of attack more than they affect lift. Think of two bodies with different masses getting the same impulse applied to them.

Thanks to Sinisa for letting me know about this discussion.

Regards, Alexander. http://staticlineinteractive.com

[format 0]

sjc

In a steady state flight, when nothing changes, there should be no affect of wind speed on descent rate.

So, how come I've timed ~10% difference (with a stopwatch) in flight_time with your Canopy Glide Simulator between 'No wind' and '13mph wind'?
(steady 'no input'; 'half input'; full input' flight)

[billvon 2,379]

>how a bullet fired horizontally and a bullet dropped in the moving stream not the
>same thing? the result is the same.. it moves vertically slightly slower than bullet
>dropped in still medium

Just as your parachute would fly differently if it were flying at 700 mph vs. 0 mph airspeed.

However, if you drop a bullet in zero speed air (standing on the ground) vs. drop a bullet in 500mph air (standing inside a 747 in flight) they will fall the same way in the same time.

[sky12345 0]

we're talking about different things.. i didnt mean the quantitative difference between an object moving at 700mph and 20mph.. i meant qualitatative diff

ok,,, one last try, more succinctly... two da vinci parachutes (essentially a round parachute with fully formed shape) r released from a balloon and from an airplane, from the same altitude simultaneously. which one will land first?

my answer: the one dropped from the balloon
this is the same answer, qualitatively, as with mythbusters bullet
once u agree with that theres no reason not to agree that a round descends slightly slower through shear layers than still air


your answer?

[sjc 0]

format

So, how come I've timed ~10% difference (with a stopwatch) in flight_time with your Canopy Glide Simulator between 'No wind' and '13mph wind'?
(steady 'no input'; 'half input'; full input' flight)

The short answer is: the model is not perfect. The longer answer is, that there are multiple variables at play and the model must accommodate the steady state as well as the response to the inputs. This requires a compromise which results in parameters that produce overall behavior. By definition it is a mathematical approximation of the reality. As the model becomes better we'll see better results.

In general, the goal of the simulator is to support the discussions such as this one. I was happy to see that it was brought up in the conversation.

Regards, Alexander. http://staticlineinteractive.com

[sky12345 0]

sjc

***So, how come I've timed ~10% difference (with a stopwatch) in flight_time with your Canopy Glide Simulator between 'No wind' and '13mph wind'?
(steady 'no input'; 'half input'; full input' flight)

The short answer is: the model is not perfect. The longer answer is, that there are multiple variables at play and the model must accommodate the steady state as well as the response to the inputs. This requires a compromise which results in parameters that produce overall behavior. By definition it is a mathematical approximation of the reality. As the model becomes better we'll see better results.

thats very poor excuse
the descent rates in no wind and in steady winds MUST match 100% ***or else*** u should ditch your 'model' completely
u may not get simulation of control inputs or turns 100% right,,, nobody can,,, but not to match the descent rates is like writing a calculator program that when entered 2*2 gives 4.4. "yeah i know it must be 4 but my model is not perfect"

"new version of my calculator will be better. 2*2 will be 4.2"

[sjc 0]

sky12345

thats very poor excuse
the descent rates in no wind and in steady winds MUST match 100% ***or else*** u should ditch your 'model' completely
u may not get simulation of control inputs or turns 100% right,,, nobody can,,, but not to match the descent rates is like writing a calculator program that when entered 2*2 gives 4.4. "yeah i know it must be 4 but my model is not perfect"

"new version of my calculator will be better. 2*2 will be 4.2"

I understand your feelings, though I disagree with the logic. If you would like to discuss this further we can do it in some other venue (at least not this thread).

Regards, Alexander. http://staticlineinteractive.com

[pchapman 261]

sky12345

>Not the same thing at all. And easily proved using regular Newtonian physics.

how a bullet fired horizontally and a bullet dropped in the moving stream not the same thing?
[...]
>horizontal and vertical components of drag are no longer independent

what do u mean independent? they r just projections of a vector, how they can be independent?

Ok, although I've been snickering at some of the stuff in this thread, I'll give you a serious answer about how it works to help you and others learn this.

Let's say a heavy round ball is dropped vertically, vs. dropped from an airplane moving fast horizontally.

Drag tends to change with the square of the speed. I'll skip all the real world numbers and we'll just say that something falling with speed 1 has a drag of 1 squared, which =1. Speed 2 gives drag 4, in some system of units.

Lets say we compare the drag when both balls are falling speed 3 vertically, but one ball is still moving forwards fast, lets say speed 10.

The drag on the ball dropping vertically is 3 squared =9.

For the ball moving forward, we can't just say that the vertical speed is 3 so the drag that affects how fast it is accelerating downwards is 3.

Instead we have to look at the whole velocity vector. Speed 3 down and 10 forward = 10.44 speed by Pythagoras, at an angle 16.70 degrees below the horizon. Total drag is 10.44 squared = 109.

The vertical component of that is 109*sin(16.70 deg) = 109*.287 = 31.28.

Voila. The ball falling straight down will accelerate straight down based on a drag of 9, while the ball arcing down with a lot of horizontal velocity, will accelerate downwards (at that particular moment in its trajectory) as befits a drag of over 31. So it will be accelerating downwards a lot slower.

There's a huge difference in the drag values, even when one is looking only at the vertical direction, despite both balls moving in the vertical direction the same speed.

At slow overall speeds you might be able to make approximate calculations for the vertical speed only, just by looking at the vertical speed of the object, treating the horizontal dimension as independent. After all, we might use a skydiving table that says that after 5 seconds of fall, a jumper will have dropped roughly 366 ft. In reality, that will differ a little depending on the forward speed of the airplane (even for jumpers all on their belly and 'falling straight').

If you want the calculations to be correct, you have to do the trigonometry and not just treat horizontal and vertical movement separately.

[unkulunkulu 0]

Another perspective: imagine a no-wind day and say you're trying to land on a moving vehicle. It's obvious that your descent rate will be the same no matter whether you're landing in the direction of this vehicle moving or in the opposite direction (let's assume that the vehicle is huge and you can easily land on it in those conditions).

Now the only thing left to understand is that the ground is your vehicle moving in the opposite direction to the wind, because you know, all movement is relative and air moving wrt earth is the same as earth moving wrt to the air.

Does this make sense?

[sky12345 0]

***THANK U***

this is a perfect explanation
even an idiot like me understood
if only all physics professors were like u!

[sky12345 0]

sjc

***
thats very poor excuse
the descent rates in no wind and in steady winds MUST match 100% ***or else*** u should ditch your 'model' completely
u may not get simulation of control inputs or turns 100% right,,, nobody can,,, but not to match the descent rates is like writing a calculator program that when entered 2*2 gives 4.4. "yeah i know it must be 4 but my model is not perfect"

"new version of my calculator will be better. 2*2 will be 4.2"

I understand your feelings, though I disagree with the logic. If you would like to discuss this further we can do it in some other venue (at least not this thread).

my logic is that its not ok to sell bullshit
your prog fails an elementary bs check
u disagree? its ok 4u to sell bs?!

[kallend 1,619]

sky12345

>Not the same thing at all. And easily proved using regular Newtonian physics.

how a bullet fired horizontally and a bullet dropped in the moving stream not the same thing? the result is the same.. it moves vertically slightly slower than bullet dropped in still medium

>horizontal and vertical components of drag are no longer independent

what do u mean independent? they r just projections of a vector, how they can be independent?

IF drag were a linear function of velocity then the horizontal component of drag would be independent of the vertical component of velocity and the vertical component of drag would be independent of the horizontal component of velocity.

However, drag is NOT a linear function of velocity, so the horizontal component of velocity does have an effect on the vertical component of drag and the vertical component of velocity does have an effect on the horizontal component of drag.

Quote

>
i had F- in physics

Why am I not surprised.

...

The only sure way to survive a canopy collision is not to have one.

[unkulunkulu 0]

Quote

However, drag is NOT a linear function of velocity, so the horizontal component of velocity does have an effect on the vertical component of drag and the vertical component of velocity does have an effect on the horizontal component of drag.

Soo, this is actually the dominating reason why trackers stay longer in the air, right? Not "lift"? Also, this must work for larger canopies (most canopies?)

And in general, what reading do you suggest on canopy aerodynamics?

[pchapman 261]

unkulunkulu

Quote

However, drag is NOT a linear function of velocity,[...]

Soo, this is actually the dominating reason why trackers stay longer in the air, right?

No. Trackers or inclined barn doors or airplane wings all still work due to a combination of lift (defined as the force generated perpendicular to the line of flight) and drag (defined as the force generated parallel to the line of flight).

This principle about 'not being able to treat vertical motion while ignoring horizontal motion' is about aerodynamic forces in general.

It applies equally well to a round cannonball that is generating no lift at all. If it is zooming forward while dropping a little, with all its speed it is generating a lot of drag, which does indeed slow down its acceleration vertically.

(Which if you don't understand the whole picture, is like some kind of magic anti-gravity lift. But it isn't lift since there's no force perpendicular to the line of flight.)

[kallend 1,619]

unkulunkulu

Quote

However, drag is NOT a linear function of velocity, so the horizontal component of velocity does have an effect on the vertical component of drag and the vertical component of velocity does have an effect on the horizontal component of drag.

Soo, this is actually the dominating reason why trackers stay longer in the air, right? Not "lift"? Also, this must work for larger canopies (most canopies?)

No, but since Mr. Chapman has already answered correctly I'll refrain from duplication.

...

The only sure way to survive a canopy collision is not to have one.

[sky12345 0]

>the horizontal component of velocity does have an effect on the vertical component of drag and the vertical component of velocity does have an effect on the horizontal component of drag.

nonsense,, tautology,, pseudoscience....your pick
its like saying that if we have a triangle with sides a b c and we blow it up 2x the increase of length of a has an effect on length of c!
thats bs,, no causality whatsoever,, the reason c became 2x is not because a or b became 2x, its because we blew the whole triangle up 2x!
same here,, as pchapman perfectly explained, greater velocity = greater drag (in quadratic proportion), and its vertical component is greater than when the object is dropped with no horizontal speed
total velocity is the primary thing, the cause of the effect,, not some mysterious crossbreeding btw horiz and vert components lol
components r arbitrary imaginary things
velocity is a real one

u saying that horiz component of speed has "an effect" on vert comp of drag is like saying that the growing child became taller because the height from crotch to floor increased! lol

[kallend 1,619]

Your F- in physics is showing, Mr. Anonymous Troll.

This has all been discussed in DZ.COM a decade ago, as I have repeatedly stated:

www.dropzone.com/cgi-bin/forum/gforum.cgi?post=1158261#1158261

(In fact Newton was aware of it in 1676; he gives an example in Principia).

...

The only sure way to survive a canopy collision is not to have one.