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I need serious help with this guy

jaydeehess said:
take any object and set
mg=0.5CDpv2A and solve for v to find terminal velocity
v=sqrt(2mg/CDpA)

I challenge anyone without a math or physics degree to now solve the time to reach terminal velocity. It will involve some calculus(integration).
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Less than one year. What do I win?
 
I must say I was delighted to study Mark Robert's work and wish to suggest the analogy that he is indeed a modern-day 'Sherlock Holmes'! Do you all agree?

Do not post off-topic and do not spam the forum with the same message over and over.
Replying to this modbox in thread will be off topic  Posted By: Lisa Simpson
 
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Congruentruth said:
I must say I was delighted to study Mark Robert's work and wish to suggest the analogy that he is indeed a modern-day 'Sherlock Holmes'! Do you all agree?
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I'm not sure how your question is on topic to this thread, but no, I don't think he's a modern-day "Sherlock Holmes".
 
triforcharity said:
Is it 30.5 seconds??
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I don't know. I am asking for the general equation for finding the time to terminal velocity of any object.

However we can calculate what the terminal velocity of a steel box column section is.

Let's take some dimensions
9 meters(30 feet) long with 4 equal sides 0.3 meters (1 foot) wide and using 0.025 meter (1 inch) thick

The mass of steel of this column is 2090 Kg
If it falls flat side down all the way this would be the slowest terminal velocity (and thus the fastest time to reach terminal velocity) and the area facing the air flow would be 2.7 square meters

The density of air is 1.2 Kg/cubic meter

The coefficient of drag of a cube (a rectilinear shape would be similar) is 1.05

so if mg=CdpAv2we have
sqrt{2090(9.8)/1.05(1.2)(2.7)}=v =77.5 meters per second or 279 kilometers per hour

In order to reach that velocity in a vaccum(no air drag) it would be falling for 8 seconds so time to terminal velocity would be greater than 8 seconds when falling in air. (because its not accellerating as fast as in a vacuum)
 
jaydeehess said:
I don't know. I am asking for the general equation for finding the time to terminal velocity of any object.

However we can calculate what the terminal velocity of a steel box column section is.

Let's take some dimensions
9 meters(30 feet) long with 4 equal sides 0.3 meters (1 foot) wide and using 0.025 meter (1 inch) thick

The mass of steel of this column is 2090 Kg
If it falls flat side down all the way this would be the slowest terminal velocity (and thus the fastest time to reach terminal velocity) and the area facing the air flow would be 2.7 square meters

The density of air is 1.2 Kg/cubic meter

The coefficient of drag of a cube (a rectilinear shape would be similar) is 1.05
so if mg=CdpAv2we have sqrt{2090(9.8)/1.05(1.2)(2.7)}=v =77.5 meters per second or 279 kilometers per hour
In order to reach that velocity in a vaccum(no air drag) it would be falling for 8 seconds so time to terminal velocity would be greater than 8 seconds when falling in air. (because its not accellerating as fast as in a vacuum)
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The highlighted parts sound like Chineese to me, as I do not speak ANY forum of math beyond algebra, except for a few medical equations.

Plus, I was just pulling a random number out of thin air. (Is there such thing as THICK air??)

I am not a math guy, but I think there might be too many variables to accurately find a TV time, but I could be wrong. I slept through most of my high school math classes.
 
jaydeehess said:
I don't know. I am asking for the general equation for finding the time to terminal velocity of any object.
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It's infinite. The velocity approaches the terminal velocity asymptotically.

The better way to think about it is to find the characteristic time of the system -- similar to a "half life" -- and argue that, for periods several times the characteristic time, the velocity is close enough to terminal as makes no practical difference.

That characteristic time is equal to the terminal velocity divided by the acceleration of gravity, which in your case is (77.5 m/s)/(10 m/s2) = 7.75 seconds.

So after, say, 15 seconds, you are basically at terminal velocity.

See here for a quick derivation.
 
R.Mackey said:
It's infinite. The velocity approaches the terminal velocity asymptotically.

The better way to think about it is to find the characteristic time of the system -- similar to a "half life" -- and argue that, for periods several times the characteristic time, the velocity is close enough to terminal as makes no practical difference.

That characteristic time is equal to the terminal velocity divided by the acceleration of gravity, which in your case is (77.5 m/s)/(10 m/s2) = 7.75 seconds.

So after, say, 15 seconds, you are basically at terminal velocity.

See here for a quick derivation.
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That is the best explanation I have seen so far. There would be then a similar formula to that of half life?
or charge/discharge of a capacitive circuit?

Several times the characteristic unit is familiar to me. In electronic, and in the charge or discharge of a capacitive cct one designs cct using the idea of 5 or 10 times. Its been 20 years though since I was expected to do any such calculations. Now, in the real world, if it don't work, we do not go about re-engineering it, we just fix the &^%$ thing.:D
 
triforcharity said:
The coefficient of drag of a cube (a rectilinear shape would be similar) is 1.05 so if mg=CdpAv[/sup]2[/sup] we have ..........v =77.5 meters per second or 279 kilometers per hour
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The highlighted parts sound like Chineese to me, as I do not speak ANY forum of math beyond algebra, except for a few medical equations.

Plus, I was just pulling a random number out of thin air. (Is there such thing as THICK air??)

I am not a math guy, but I think there might be too many variables to accurately find a TV time, but I could be wrong. I slept through most of my high school math classes.
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The coefficient of drag is a figure that relates how the specific shape affects the drag on an object. Obviously a bullet shape will flow through the air better than a flat plate even if both are presenting the same surface area to the air flow.

I skipped over a lot of detail such as the fact that "p" is the desity of air and how I determined what it was, or where I got the equation from. It is all available on the net with one's search emgine as one's friend.

However understanding what one reads would require that one does have a good grasp of algebra and how to manipulate an equation.

Sunray's buddy, being an engineer of course, should have no problem in that arena.

An aeronautical engineer would be more familiar with equations such as these whereas a civil engineer would have little, or at least only occassional use for them.
 
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jaydeehess said:
That is the best explanation I have seen so far. There would be then a similar formula to that of half life?
or charge/discharge of a capacitive circuit?
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Yup. The velocity approaches terminal velocity over time hyperbolically.

The formalism is almost the same as a capacitor charging example. Here, suppose we measure the amount of charge Q that flows through a circuit with applied ideal voltage V, resistance R, and capacitor value C. The equation for charge is as follows:

0 = Q / C + Q' R - V​

This is a simple, first-order ODE in one variable, since V is a constant. Rewrite as:

Q' = (V / R) - (Q / R C)​

And the familiar solution, where R C is the characteristic time:

Q = C V (1 - e-t/R C)​

This is similar to (but not quite) the same as our situation. For drag, we have a slightly different form -- now v is velocity:

v' = (m g) - (1/2 Cd ρ v2)​

The square term changes the equation quite a bit. Instead of a simple exponential, the solution now takes the form of a hyperbolic tangent function, and solving the equation is more painful. But qualitatively the behavior is the same -- the tanh function can be thought of as a "more square" decaying exponential, one with a more pronounced knee around the characteristic time.

jaydeehess said:
Several times the characteristic unit is familiar to me. In electronic, and in the charge or discharge of a capacitive cct one designs cct using the idea of 5 or 10 times. Its been 20 years though since I was expected to do any such calculations. Now, in the real world, if it don't work, we do not go about re-engineering it, we just fix the &^%$ thing.:D
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In that spirit, you can also just drop things and measure their effective velocity. Find yourself a nice, tall, open stairwell, a stopwatch, and a bagfull of pinecones or ping-pong balls, something draggy and safe. The transition from acceleration to reaching terminal velocity is pretty sharp, and should be directly observable with little effort.
 
Sunray Breaker said:
oh god...here we go...He's got firefighters saying they found a secondary device and that theres a bomb in the building...Just when I thought I was done with this dude.

HELP!!! I know this is BS but it's a tough one (keep in mind a recovering twoofer and am still learning critical thinking skills
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Captain Jay Swithers

I saw a large section of it blasting out, which led me to believe it was just an explosion. I thought it was a secondary device, but I knew that we had to go.

But one thing that did happen was an ambulance pulled up which was very clean. So I assumed that the vehicle had not been in the - what I thought was an explosion at the time, but was the first collapse.

http://graphics8.nytimes.com/packages/pdf/nyregion/20050812_WTC_GRAPHIC/9110172.PDF
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Sunray Breaker said:
oh god...here we go...He's got firefighters saying they found a secondary device and that theres a bomb in the building...Just when I thought I was done with this dude.

HELP!!! I know this is BS but it's a tough one (keep in mind a recovering twoofer and am still learning critical thinking skills
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If you look at the video of that statement, it was shot after at least one of the collapses. That's your first clue that it was taken out of context. The clip is actually from a bomb scare at Stuyvesant High School. See here for more details.
 
willhaven said:
Funny. CTs say that the buildings fell into their own footprint like a professional controlled demolition, yet any columns that fell well outside of the footprint of the WTC towers was obviously proof of explosive detonations.

When people use that kind of logic, you can't win.

And well more than "a few beams" fell out of the footprint. Entire buildings were crushed and had to be demolished because they were hit with debris from the WTC towers.
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My sentiments EXACTLY!!
 
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triforcharity said:
Is someone can help me out, I am having issues.

direct link

http://triforcharity.stripgenerator.com/2009/06/28/cdboth/
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cdboth.png


TaDaaaa!
 
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