Keep in mind that a bomb or artillery shell is aerodynamically streamlined (at least to some extent) and, therefore, don't have much resistance from the air.  They also have a high density (weight to volume ratio).  Human bodies are not very streamlined and have relatively low density.  Consequently, air resistance will have much greater impact on the human body than on a bomb or projectile.

The 1200/1250 feet "forward throw" discussed above applies only to human bodies and will vary between humans based on their weight and the body volume that is available to provide air resistance.

Where did the 1200/1250 number come from, not disputing the number just trying to understand the context. Is it based on the Cooper metrics.

You are not allowed to view links. Register or Login

You are not allowed to view links. Register or Login

Keep in mind that a bomb or artillery shell is aerodynamically streamlined (at least to some extent) and, therefore, don't have much resistance from the air.  They also have a high density (weight to volume ratio).  Human bodies are not very streamlined and have relatively low density.  Consequently, air resistance will have much greater impact on the human body than on a bomb or projectile.

The 1200/1250 feet "forward throw" discussed above applies only to human bodies and will vary between humans based on their weight and the body volume that is available to provide air resistance.

Where did the 1200/1250 number come from, not disputing the number just trying to understand the context. Is is based on the Cooper metrics.

Those numbers are basically just an estimate based on the experience of skydivers.  I think there are charts on this subject in some skydiver type books.  Cooper's speed and altitude (about 225 MPH true airspeed and 10,000 feet above sea level) are within the general range of experienced skydivers.  I imagine that present day skydivers, such as 377, who probably use a Twin Otter to get to about 12,500 feet, will jump at a speed of about 150 MPH true airspeed.  Of course, if 377 jumps from a DC-9 at 20,000 feet and doing 300 MPH, he will travel further down the airliner's track before his forward motion is completely stopped.

If Cooper ended up in a head down descent, he would probably be doing 180+ MPH which is lower than the 225 MPH that he was doing when he jumped.  He would slow down from the 225 MPH to his terminal velocity of 180+ real fast.

Terminal velocity can be 200mph in a ball/head down to 120mph spread belly, big range, the 1200 ft seems way too short for Cooper metrics but 1.4m too far with resistance.. I have to do more research, I always assumed "normal" exit speeds were 80-90mph, a 1200 ft sounds more right there with a belly drop. But at even at a half mile (2500 ft) that means that the flightpath would've had to briefly cross the I5 over Hayden..

Flyjack, Let's start over.

First, a hypothetical illustration.  Assume an aircraft flying two miles above sea level, no surface winds or winds aloft, and a ground speed of 240 MPH (four miles per minute).  Let's add a further simplification, there is NO ATMOSPHERE as well (repeat, this is a HYPOTHETICAL illustration).

An object (any object) that is dropped from this aircraft under these conditions will, with respect to the aircraft, stay directly under the aircraft until it strikes the earth.  While descending from the aircraft, this object is only subjected to the laws of gravity which are explained in elementary physics textbooks, Wikipedia, and other online Internet sources.  Using the relatively simple equations for a body falling from rest in a vacuum, the time of the fall of the object from the aircraft to the ground can be calculated. Then the distance that the aircraft travelled in that time can also be calculated.

The weight vector (as used here, vector means a force acting in a specified direction) of any body, including the one described above, is ALWAYS pointing towards the center of the earth.

Now let's revisit the above problem and remove the no atmosphere assumption.  Before considering horizontal motion, let's assume that an object is dropped from a stationary balloon.  We now have to consider the atmosphere's impact (or air resistance) on the falling body.  For purely vertical motion, this is also relatively simple and problems along this line are considered in elementary physics books and such.  Other things being the same, it would take the same object, falling from the same height, longer to reach the ground with air resistance than if it was in a vacuum.

If the falling object has sufficient altitude, it will reach a speed where the air resistance is equal to the weight of the object.  The air resistance, or drag force, in this instance is the same magnitude and pointing in the opposite direction of the weight.  In addition, the drag force is ALWAYS parallel to and pointing in the opposite direction of the free-stream velocity vector.  And in this example, the free-stream velocity vector also points straight down or towards the center of the earth.

When we add horizontal motion in an atmosphere to our problem, the situation goes from relatively simple to intractable.  The equations governing this horizontal motion become about fourth-order quadratic non-linear equations and require something on the order of a Cray supercomputer to solve.  That is why we don't see tables of actual computations.  Instead we have estimates and guesses about the horizontal motion of the falling body.

Incidentally, the term "throw forward" is not an aerodynamics term but a term used by skydivers.  Page 233, The Skydiver's Handbook, 10th Revised Edition, by Dan Poynter and Mike Turoff, contains a discussion on forward throw and includes a chart of speeds and distances.  The last thing before the chart is the phrase "we're not going to delve into the physics of deceleration resistance here".  That says it all.

How about taking a look at the online information concerning free falling bodies and see if that can explain things a bit better than me.  If it doesn't, feel free to ask more questions.

Robert99     

You are not allowed to view links. Register or Login

Flyjack, Let's start over.

First, a hypothetical illustration.  Assume an aircraft flying two miles above sea level, no surface winds or winds aloft, and a ground speed of 240 MPH (four miles per minute).  Let's add a further simplification, there is NO ATMOSPHERE as well (repeat, this is a HYPOTHETICAL illustration).

An object (any object) that is dropped from this aircraft under these conditions will, with respect to the aircraft, stay directly under the aircraft until it strikes the earth.  While descending from the aircraft, this object is only subjected to the laws of gravity which are explained in elementary physics textbooks, Wikipedia, and other online Internet sources.  Using the relatively simple equations for a body falling from rest in a vacuum, the time of the fall of the object from the aircraft to the ground can be calculated. Then the distance that the aircraft travelled in that time can also be calculated.

The weight vector (as used here, vector means a force acting in a specified direction) of any body, including the one described above, is ALWAYS pointing towards the center of the earth.

Now let's revisit the above problem and remove the no atmosphere assumption.  Before considering horizontal motion, let's assume that an object is dropped from a stationary balloon.  We now have to consider the atmosphere's impact (or air resistance) on the falling body.  For purely vertical motion, this is also relatively simple and problems along this line are considered in elementary physics books and such.  Other things being the same, it would take the same object, falling from the same height, longer to reach the ground with air resistance than if it was in a vacuum.

If the falling object has sufficient altitude, it will reach a speed where the air resistance is equal to the weight of the object.  The air resistance, or drag force, in this instance is the same magnitude and pointing in the opposite direction of the weight.  In addition, the drag force is ALWAYS parallel to and pointing in the opposite direction of the free-stream velocity vector.  And in this example, the free-stream velocity vector also points straight down or towards the center of the earth.

When we add horizontal motion in an atmosphere to our problem, the situation goes from relatively simple to intractable.  The equations governing this horizontal motion become about fourth-order quadratic non-linear equations and require something on the order of a Cray supercomputer to solve.  That is why we don't see tables of actual computations.  Instead we have estimates and guesses about the horizontal motion of the falling body.

Incidentally, the term "throw forward" is not an aerodynamics term but a term used by skydivers.  Page 233, The Skydiver's Handbook, 10th Revised Edition, by Dan Poynter and Mike Turoff, contains a discussion on forward throw and includes a chart of speeds and distances.  The last thing before the chart is the phrase "we're not going to delve into the physics of deceleration resistance here".  That says it all.

How about taking a look at the online information concerning free falling bodies and see if that can explain things a bit better than me.  If it doesn't, feel free to ask more questions.

Robert99   

Thanks Robert99,

I do understand how complex this is,, to calculate.. I have read lots of info and I get the concepts, a little unsure of the maths, to estimate. The difference between dropping a 200lb bomb and a 200lb body is drag.. the bomb calculators give about 1.4 miles.. after 26 seconds

I am trying to understand if it is even possible for him to reach Smith Lake in a no pull.

For sake of reference,

How far horizontally (ground distance) from the exit point would a 200lb body land from a 10000 ft drop at 200 mph speed assuming a head/feet first with a terminal velocity of 200mph?

Will you please identify a place on a map, or some flight path, where you want him to jump from?

 

You are not allowed to view links. Register or Login

Will you please identify a place on a map, or some flight path, where you want him to jump from?

 

The I5 on Hayden Is...  I realize this flightpath is disputed, this isn't the issue right now, but it shows the flightpath almost over the I5 about Hayden and follow it for a bit...

I get the aerodynamic theories, I don't get a confident estimate, the maths. If the body can only travel .25 miles it is a stretch, if it is closer to 1.25 miles then Smith Lake is reachable.. I realize it is an estimate and there are variables.

I have read that a head/feet down could have a Terminal Velocity up to 200mph, close to exit speed. In a best case scenario for distance, he could have left the plane at essentially TV in a head/feet down no pull near I5 on Hayden. To reach Smith Lake, it is about 1.25 miles.. are we there. close or way way short. This isn't an argument over the flightpath, assume the FBI black line IS accurate..,

bomb trajectory hits land at 1.4 mi, very low drag,,  a body at TV from exit in a head/feet no pull??

closer -

You are not allowed to view links. Register or Login

You are not allowed to view links. Register or Login

Will you please identify a place on a map, or some flight path, where you want him to jump from?

 

The I5 on Hayden Is...  I realize this flightpath is disputed, this isn't the issue right now, but it shows the flightpath almost over the I5 about Hayden and follow it for a bit...

I get the aerodynamic theories, I don't get a confident estimate, the maths. If the body can only travel .25 miles it is a stretch, if it is closer to 1.25 miles then Smith Lake is reachable.. I realize it is an estimate and there are variables.

I have read that a head/feet down could have a Terminal Velocity up to 200mph, close to exit speed. In a best case scenario for distance, he could have left the plane at essentially TV in a head/feet down no pull near I5 on Hayden. To reach Smith Lake, it is about 1.25 miles.. are we there. close or way way short. This isn't an argument over the flightpath, assume the FBI black line IS accurate..,

bomb trajectory hits land at 1.4 mi, very low drag,,  a body at TV from exit in a head/feet no pull??

Flyjack, Let me say again that there is NO SIMPLE WAY to compute the horizontal distance that the body would travel after separating from the aircraft.  And again, that is why "rules of thumb", estimates, guesses, etc., are used in this instance.

Let's discuss "terminal velocity" a bit.  In the example we have been discussing, terminal velocity is when the drag force on the object equals the weight of that object.  This means that at higher altitudes, the terminal velocity will be greater than the terminal velocity at sea level.  The values of 120 MPH for a skydiver in a stable spread and 180+ MPH for head first descent by the same skydiver apply only to sea level.  At higher altitudes, the velocities will be greater.  But for both the 120 MPH and 180+ MPH situations just discussed, the drag on the skydiver will be the same, and that is equal to his weight.

The same principle applies to aircraft.  With a given aircraft configuration, the drag on an aircraft will be the same at 10,000 feet as it is at sea level for a given Indicated Airspeed.  However, the aircraft will have a higher True Airspeed at 10,000 feet.  At sea level, and with no errors in the airspeed instrument or its pressure measuring system ("position error"), Indicated Airspeed and True Airspeed are the same.

Flyjack, it would be helpful if you would give me some idea of your technical background, both in mathematics and aerodynamics.

Robert99

will get some numbers previously calculated - in meantime:

'snowmman

Jul 22, 2008, 3:20 PM
Post #3718 of 56871 (51381 views)
Shortcut
   
getting into the Columbia [In reply to]    Can't Post
In reply to:
And no DZ defined to date accomodates any of this!

at 20:17, there's probably enough forward throw for a no-pull to have Cooper in the Columbia. (from the FBI flight path)

And if you subtract the apparent error minute, that's 20:16

I think the data suggesting LZ sooner before the Columbia is pretty weak and shouldn't override the more likely Columbia LZ (based on the money)

Also, since Columbia/PDX area is an easier visual spot, compared to LZ before that, it makes sense?

Assuming Cooper is not a skydiver, the increased risk of drowning might not have registered.'

You are not allowed to view links. Register or Login

You are not allowed to view links. Register or Login

You are not allowed to view links. Register or Login

Will you please identify a place on a map, or some flight path, where you want him to jump from?

 

The I5 on Hayden Is...  I realize this flightpath is disputed, this isn't the issue right now, but it shows the flightpath almost over the I5 about Hayden and follow it for a bit...

I get the aerodynamic theories, I don't get a confident estimate, the maths. If the body can only travel .25 miles it is a stretch, if it is closer to 1.25 miles then Smith Lake is reachable.. I realize it is an estimate and there are variables.

I have read that a head/feet down could have a Terminal Velocity up to 200mph, close to exit speed. In a best case scenario for distance, he could have left the plane at essentially TV in a head/feet down no pull near I5 on Hayden. To reach Smith Lake, it is about 1.25 miles.. are we there. close or way way short. This isn't an argument over the flightpath, assume the FBI black line IS accurate..,

bomb trajectory hits land at 1.4 mi, very low drag,,  a body at TV from exit in a head/feet no pull??

Flyjack, Let me say again that there is NO SIMPLE WAY to compute the horizontal distance that the body would travel after separating from the aircraft.  And again, that is why "rules of thumb", estimates, guesses, etc., are used in this instance.

Let's discuss "terminal velocity" a bit.  In the example we have been discussing, terminal velocity is when the drag force on the object equals the weight of that object.  This means that at higher altitudes, the terminal velocity will be greater than the terminal velocity at sea level.  The values of 120 MPH for a skydiver in a stable spread and 180+ MPH for head first descent by the same skydiver apply only to sea level.  At higher altitudes, the velocities will be greater.  But for both the 120 MPH and 180+ MPH situations just discussed, the drag on the skydiver will be the same, and that is equal to his weight.

The same principle applies to aircraft.  With a given aircraft configuration, the drag on an aircraft will be the same at 10,000 feet as it is at sea level for a given Indicated Airspeed.  However, the aircraft will have a higher True Airspeed at 10,000 feet.  At sea level, and with no errors in the airspeed instrument or its pressure measuring system ("position error"), Indicated Airspeed and True Airspeed are the same.

Flyjack, it would be helpful if you would give me some idea of your technical background, both in mathematics and aerodynamics.

Robert99

will get some numbers previously calculated - in meantime:

'snowmman

Jul 22, 2008, 3:20 PM
Post #3718 of 56871 (51381 views)
Shortcut
   
getting into the Columbia [In reply to]    Can't Post
In reply to:
And no DZ defined to date accomodates any of this!

at 20:17, there's probably enough forward throw for a no-pull to have Cooper in the Columbia. (from the FBI flight path)

And if you subtract the apparent error minute, that's 20:16

I think the data suggesting LZ sooner before the Columbia is pretty weak and shouldn't override the more likely Columbia LZ (based on the money)

Also, since Columbia/PDX area is an easier visual spot, compared to LZ before that, it makes sense?

Assuming Cooper is not a skydiver, the increased risk of drowning might not have registered.'

If the airliner's position was at the specified DME distance south of the Battleground VORTAC at 8:18 PM, then this moves the jump time to about 8:12, as previously estimated by some, and it probably could not have been later than about 8:14 at most in order for the money to ever make it to Tina Bar.