[Justin Aerial]

I have an interesting article on my site with the relative heights of various TV transmitters all next to each other.
As an ammusing addition I calculated how long it`d take an object to hit the ground if dropped from the top of each of these TXs, the speed it`d be going when it hit the ground, and the number of asprin required to alleviate the pain if it hit you.
No not the last bit (though it`d be interesting none the less).
The trouble is that these calculations were done using the formula for free space (i.e. a vacuum).
So the question is what are the correct figures for air, for something weighty and dense, like a 19mm ring spanner. The latter weighs 138g, and is 225mm x 20mm x 5mm (all averages).

[grahamlthompson]

The weight is not relevant. It's the wind resistance that is, according to the air pressure every object will have a terminal velocity which once reached won't be exceeded.(once the retarding force matches the accelerating one (nominally 1G). If I remember right for an average human it's around 160mph.

[Justin Aerial]

Originally Posted by grahamlthompson

The weight is not relevant. It's the wind resistance that is, according to the air pressure every object will have a terminal velocity which once reached won't be exceeded.(once the retarding force matches the accelerating one (nominally 1G). If I remember right for an average human it's around 160mph.

Technically the weight might not be relevant, but the density most assuredly is. What you`ve got is the force of gravity accelerating the object on the one hand, and the air resistance (drag) slowing it down on the other.
My contention is that a sphere made of lead would have a higher terminal velocity than one made of aluminium.
If you have something dense the effect of drag will be relatively less, and as that object gets bigger the surface area to volume ratio changes as well doesn`t it, so doesn`t that also have an effect ?

[Mark C]

Originally Posted by grahamlthompson

The weight is not relevant. It's the wind resistance that is, according to the air pressure every object will have a terminal velocity which once reached won't be exceeded.(once the retarding force matches the accelerating one (nominally 1G). If I remember right for an average human it's around 160mph.

A bit lower, about 125 mph, (still more than enough to be a 'shovel job' !)

[grahamlthompson]

Originally Posted by Justin Aerial

Technically the weight might not be relevant, but the density most assuredly is. What you`ve got is the force of gravity accelerating the object on the one hand, and the air resistance (drag) slowing it down on the other.
My contention is that a sphere made of lead would have a higher terminal velocity than one made of aluminium.
If you have something dense the effect of drag will be relatively less, and as that object gets bigger the surface area to volume ratio changes as well doesn`t it, so doesn`t that also have an effect ?

The density is not relevant either. It's the old adage which falls faster in a vacuum a pound of feathers or a pound of lead. The answer is they both accelerate the same having an acceleration of 32.2ft/sec/sec After two seconds both will have reached a velocity of 64.4ft/sec.

A hollow aluminium sphere with the same surface drag and dimensions as a solid lead one will accelerate at the same rate (On Earth about 32.2ft/sec squared).

http://www.haverford.edu/educ/knight...ccelarator.htm

If you want to find the depth of a hole (ignoring wind resistance) Time a round pebble till you hear it hit the bottom. Say it takes 20 seconds. Because the acceleration is linear after 10 seconds the pebble will be travelling at it's average velocity. In this case 322ft/sec. An object travelling at 322 ft/sec for 20 seconds will have travelled 6440ft (some hole )

[rjb101]

Some Italian bloke did some stuff on this in the Renaissance. Chucking different sizes of canon balls off the Leaning Tower of Pisa and noting they arrived at the bottom at the same time.


The wife has just piped up and suggested it was "Whats his name" Thank-you dear

Edit. Galileo. Thank you Google

[grahamlthompson]

Originally Posted by rjb101

Some Italian bloke did some stuff on this in the Renaissance. Chucking different sizes of canon balls off the Leaning Tower of Pisa and noting they arrived at the bottom at the same time.


The wife has just piped up and suggested it was "Whats his name" Thank-you dear

Edit. Galileo. Thank you Google

I bet it took a lot less than 20 seconds

[Analogue110]

This experiment was also carried out on the moon, using a hammer and a feather. I cannot remember which moon landing it was.

[grahamlthompson]

Originally Posted by Analogue110

This experiment was also carried out on the moon, using a hammer and a feather. I cannot remember which moon landing it was.

Well remembered . As always Google to the rescue

http://www.youtube.com/watch?v=5C5_dOEyAfk

Apollo 15 by the way

[chrisy]

Originally Posted by Analogue110

This experiment was also carried out on the moon, using a hammer and a feather. I cannot remember which moon landing it was.

Of course, on Earth the feather would fly away on the wind

[maldonian]

Originally Posted by grahamlthompson

The weight is not relevant. It's the wind resistance that is, according to the air pressure every object will have a terminal velocity which once reached won't be exceeded.(once the retarding force matches the accelerating one (nominally 1G). If I remember right for an average human it's around 160mph.

The weight is relevant. The downward force is the weight, the upward force is the air resistance or drag. At terminal velocity the drag is equal to the weight. The drag is proportional to the projected area of the object in the direction its moving, the square of its velocity, and the density of the air. Terminal velocity increases with size for similarly shaped objects of the same density (because the weight is proportional to the volume and the drag is proportional to the projected area).

The terminal velocity of a human depends on the orientation of the body. Skydivers can free fall much faster when diving head first with their arms by their sides and legs together than in the usual spread eagled position with their front facing the ground. Opening a parachute reduces the terminal velocity to a safe value by greatly increasing the drag.

The question is, would a large spanner get anywhere near it's terminal velocity if dropped from the top of Emley Moor's mast, or would it still be accelerating at anywhere near 32 ft per sec per sec just before it reaches the ground?

The density is not relevant either. It's the old adage which falls faster in a vacuum a pound of feathers or a pound of lead.

Emley Moor isn't in a vacuum.

Originally Posted by chrisy

Of course, on Earth the feather would fly away on the wind

Indeed.

[Justin Aerial]

Originally Posted by Justin Aerial

I have an interesting article on my site with the relative heights of various TV transmitters all next to each other.
As an ammusing addition I calculated how long it`d take an object to hit the ground if dropped from the top of each of these TXs, the speed it`d be going when it hit the ground, and the number of asprin required to alleviate the pain if it hit you.
No not the last bit (though it`d be interesting none the less).
The trouble is that these calculations were done using the formula for free space (i.e. a vacuum).
So the question is what are the correct figures for air, for something weighty and dense, like a 19mm ring spanner. The latter weighs 138g, and is 225mm x 20mm x 5mm (all averages).

So, come on all you brain boxes out there, Digital Spy`s reputation is on the line here. What is the correct time to hit the ground, and speed when doing so, for the aforementioned 19mm combination spanner, dropped from the top of the TXs in (above linked) the article on my site ?

[Nigel Goodwin]

Originally Posted by Justin Aerial

So, come on all you brain boxes out there, Digital Spy`s reputation is on the line here. What is the correct time to hit the ground, and speed when doing so, for the aforementioned 19mm combination spanner, dropped from the top of the TXs in (above linked) the article on my site ?

There are too many variables to calculate accurately, and I imagine it would vary depending on the exact way it rotated on the way done.

On a more relevent (and practical) theme, it would happily go straight through a safety helmet

This happened to a friends father, with a spanner dropped from the top of an oil drilling rig - and while it went through his safety helmet, it did slow it enough to survive - although he suffered a fractured skull.

[Justin Aerial]

Originally Posted by maldonian

The weight is relevant. The downward force is the weight, the upward force is the air resistance or drag. At terminal velocity the drag is equal to the weight. The drag is proportional to the projected area of the object in the direction its moving, the square of its velocity, and the density of the air. Terminal velocity increases with size for similarly shaped objects of the same density (because the weight is proportional to the volume and the drag is proportional to the projected area).

The terminal velocity of a human depends on the orientation of the body. Skydivers can free fall much faster when diving head first with their arms by their sides and legs together than in the usual spread eagled position with their front facing the ground. Opening a parachute reduces the terminal velocity to a safe value by greatly increasing the drag.

The question is, would a large spanner get anywhere near it's terminal velocity if dropped from the top of Emley Moor's mast, or would it still be accelerating at anywhere near 32 ft per sec per sec just before it reaches the ground?

Emley Moor isn't in a vacuum.

Indeed.

We`ve been thnking about it and we think it depends if the spanner falls "end on" or "flat". The former would obviously have less aerodynamic drag. We think that the spanner is heavier at one end, namely the open ended end. Further more I reckon the drag would be slightly greater for the ring end. So, for the purposes of this we`ll assumes the spanner falls with an equivalent surface area of 100mm (or 1cm).
We`ll ignore surface drag on the spanner, because that just makes it too complicated.

[grahamlthompson]

Originally Posted by maldonian

The weight is relevant. The downward force is the weight, the upward force is the air resistance or drag. At terminal velocity the drag is equal to the weight. The drag is proportional to the projected area of the object in the direction its moving, the square of its velocity, and the density of the air. Terminal velocity increases with size for similarly shaped objects of the same density (because the weight is proportional to the volume and the drag is proportional to the projected area).

The terminal velocity of a human depends on the orientation of the body. Skydivers can free fall much faster when diving head first with their arms by their sides and legs together than in the usual spread eagled position with their front facing the ground. Opening a parachute reduces the terminal velocity to a safe value by greatly increasing the drag.

The question is, would a large spanner get anywhere near it's terminal velocity if dropped from the top of Emley Moor's mast, or would it still be accelerating at anywhere near 32 ft per sec per sec just before it reaches the ground?



Emley Moor isn't in a vacuum.



Indeed.

The mass is only relevant in relation to the objects shape. Two objects of identical mass and density one formed into a sphere and the other into a flat plate will have totally different terminal velocities. .


Terminal Velocity

V = √((2mg)/(pAC))

where V is terminal velocity, m is mass of the falling object, g is the gravitaional acceleration (9.81 m/sec²), p is the density of the medium the object is falling through (1.269 kg/m³ for air), A is the object's cross sectional area, and C is the object's drag coefficient. Largest problem is finding the drag coefficient For a man in an upright position, it's roughly 1.0, for a man in prone position, about 1.3, and for a rough sphere, about 0.4, while a smooth sphere is about 0.1.

Can anyone find a formula for the rate of acceleration based on the above related to time. Given this it should be easy enough to produce a graph of distance travelled over time for a given object.

[grahamlthompson]

Originally Posted by Justin Aerial

We`ve been thnking about it and we think it depends if the spanner falls "end on" or "flat". The former would obviously have less aerodynamic drag. We think that the spanner is heavier at one end, namely the open ended end. Further more I reckon the drag would be slightly greater for the ring end. So, for the purposes of this we`ll assumes the spanner falls with an equivalent surface area of 100mm (or 1cm).
We`ll ignore surface drag on the spanner, because that just makes it too complicated.

posted twice in error.

[JamesE]

v=u+at, s=ut + 1/2 at*t
328m high, g is about 10m/s/s
328 = 0.5 x 10 x t x t, t=8.1
v=gt, thus v=81m/s

ignoring air resistance, which is reasonable.

[grahamlthompson]

Originally Posted by JamesE

v=u+at, s=ut + 1/2 at*t
328m high, g is about 10m/s/s
328 = 0.5 x 10 x t x t, t=8.1
v=gt, thus v=81m/s

ignoring air resistance, which is reasonable.

Ignoring air resistance I just worked out for a drop of 1000ft the spanner would be doing around 169mph at impact

81Meters/sec = 189mph.

[janet owen]

Originally Posted by grahamlthompson

Ignoring air resistance I just worked out for a drop of 1000ft the spanner would be doing around 169mph at impact

81Meters/sec = 189mph.

Delighted you gave the answer in £ S D, have a pint on me,so ill send you 1/9d

[grahamlthompson]

Originally Posted by janet owen

Delighted you gave the answer in £ S D, have a pint on me,so ill send you 1/9d

If only

[Justin Aerial]

Originally Posted by grahamlthompson

Ignoring air resistance I just worked out for a drop of 1000ft the spanner would be doing around 169mph at impact

81Meters/sec = 189mph.

When I worked it out I made it 180mph, but that was using G as 9.8 m/s/s, i.e about the same.
But are we saying that air resistance really isn`t going to have any effect, or any significant effect ?
I`d have thought it might do, esp when the velocity got up to 100mph !
Is anyone clever enough to know what effect the air resistance would have ?

[Justin Aerial]

Originally Posted by JamesE

v=u+at, s=ut + 1/2 at*t
328m high, g is about 10m/s/s
328 = 0.5 x 10 x t x t, t=8.1
v=gt, thus v=81m/s

ignoring air resistance, which is reasonable.

Ah, but is it reasonable......

[Nigel Goodwin]

Originally Posted by Justin Aerial

When I worked it out I made it 180mph, but that was using G as 9.8 m/s/s, i.e about the same.
But are we saying that air resistance really isn`t going to have any effect, or any significant effect ?
I`d have thought it might do, esp when the velocity got up to 100mph !
Is anyone clever enough to know what effect the air resistance would have ?

I always thought terminal velocity was something like 180mph?.

[Justin Aerial]

Originally Posted by Nigel Goodwin

I always thought terminal velocity was something like 180mph?.

But it varies according to the density and the drag coeffiecient of the object.

[grahamlthompson]

Originally Posted by Justin Aerial

When I worked it out I made it 180mph, but that was using G as 9.8 m/s/s, i.e about the same.
But are we saying that air resistance really isn`t going to have any effect, or any significant effect ?
I`d have thought it might do, esp when the velocity got up to 100mph !
Is anyone clever enough to know what effect the air resistance would have ?

On a flat road and with zero wind the top speed of your car is determined by available power and the combined effects of friction and wind resistance. Hang a spanner on the outside I doubt whether the difference would be noticeable. It takes a pretty powerfull car to produce an acceleration of 1G. Mass has little or no effect on this only in so far as it increase the friction element so mass affects the rate at which the vehicle accelerates rather than top speed. Compare the power to weight ratio of a HS125 train with a car capable of a similar top speed.

What we already know given the assumptions are correct at around 180mph or so the spanner will stop accelerating, but only a small percentage of the flight time will at anything near this velocity. It will obviously make some difference the better the streamlining the less the effect. The perfect shape for a falling object is a rain drop as it automatically takes up the shape giving minimum resistance.