By continuing to use this site, you agree to our use of cookies. Find out more
Forum sponsored by:
Forum sponsored by CML

Pilot Bails Out At Low Level

All Topics | Latest Posts

Search for:  in Thread Title in  
PETER BRUCE - Eastchurch Gap18/05/2017 21:55:16
avatar
698 forum posts
104 photos

Looks like I could have been out by 100%embarrassed = Ouch

PETER BRUCE - Eastchurch Gap18/05/2017 22:49:47
avatar
698 forum posts
104 photos

Perhaps I was so wrong with my height estimate because it's a big model! dont know

Martin Harris18/05/2017 23:09:30
avatar
6995 forum posts
171 photos

I fly my >7kg models with an altimeter - we have a 400' height limit for them and it's just enough to do a decent loop...

Mike Etheridge 119/05/2017 00:39:07
1256 forum posts
312 photos

So height = 100 feet (S)

initial Velocity = 27 MPH = 36 Feet per second ( U)

Using the equation V Squared = U squared + 2 F S

Then V Squared = 36 squared + 2x32x100

V Squared = 1296+6400

V = 87.72 Feet per second.

Using the equation V=U +FT

Then T = 87.72-36/32

T = 1.61 Seconds

First time I have done one of these calculations since 1965---the height could be wrong Pete and hence this calculation!

PETER BRUCE - Eastchurch Gap19/05/2017 08:22:49
avatar
698 forum posts
104 photos

Hi Mike. WoW. Looks like you cracked it and I crapped it with my estimated height. Does this mean I will now be scolded Sundayindecision. Think me needs an altimeter...

The Wright Stuff19/05/2017 14:25:10
avatar
908 forum posts
198 photos
Posted by Martin Harris on 18/05/2017 20:36:23:

Any proper physicists out there?

Yes, but I'm afraid I'm late to the party!

Yes, the horizontal velocity is irrelevant for the purposes of the calculation. It's possible it influences the downward acceleration if it produces any lift (after all, that is how a plane flies) but for a tumbling, free-falling camera, we can probably ignore that.

s = u*t + 1/2*g*t^2 is the correct formula to use for constant acceleration, as others have said. This ignores air resistance. If we use u = 0 (zero initial vertical speed) and g=10 (we are not too interested in precision) then we have:

s = 5*t^2, where s is the height in metres and t is the falling time in seconds. So if the falling time was 7 seconds (I don't know if the video plays in real time - I measured closer to 9 seconds on my computer) then we have a height of 245 metres, or 800 feet, as Martin said above.

HOWEVER, air resistance is very much not negligible, and I suspect the camera was close to terminal velocity for a large proportion of the falling time. All we have done here is calculate an upper limit. It was probably MUCH lower.

If we had assumed that it fell under constant acceleration, then it would have hit the ground at v = g*t = 10*7 = 70 m/s, or 157 mph. I suspect there would have not been much left of the camera if this had indeed been the case.

It's very possible to calculate terminal velocities - at least approximately - since we know the density of the air, and the measurements of the camera. What is also needed is the weight (yes, yes, mass) of the camera. Peter, if you have this, I will attempt to calculate the terminal velocity and hence the original height...

Edited By The Wright Stuff on 19/05/2017 14:37:33

TIM Shaw19/05/2017 14:43:23
162 forum posts
41 photos

In an ideal world, the basic Physics is actually very simple.

Ignoring drag, an object accelerates at 9.81 m/s/s, or 32 fps/s

The video timer gives us the 7 second fall time.

An object falling from a height covers distance, S (when I was at school) which is equal to UT + 1/2 x a x t squared.

Since you were flying straight and level, the initial vertical component (or velocity) is zero, hence the first term is also zero, so the distance covered is simply 1/2 x 9.81 x 7 x 7

Which is 240.3m, or 788 feet..

If this were the case, your camera would have hit the ground travelling at a vertical speed given by v = 0 +at, which I think is 69 m/s, or 225fps - about half the muzzle velocity of a non FAC air rifle, and I doubt it would be in the condition it appears to be....

However, in the real world, objects achieve a terminal velocity, at which point the resistance to falling due to aerodynamic effects, primarily drag, equals the accelerative force of gravity, and the object then falls at constant velocity.

So the really interesting question here is just how much drag your camera has.

If you really were 100 feet up, then it only achieved an average speed of about 14 fps, about a tenth of that under ideal conditions.

I guess this really only goes to show how far from ideal our real world actually is....

The Wright Stuff19/05/2017 14:47:36
avatar
908 forum posts
198 photos

Tim, I assume our posts crossed! Great minds think alike!

The Wright Stuff19/05/2017 14:51:25
avatar
908 forum posts
198 photos
Posted by TIM Shaw on 19/05/2017 14:43:23:

I guess this really only goes to show how far from ideal our real world actually is....

I actually used to give a talk on 'the art of approximation' using exactly this concept as an example (in my case it was the time taken for a pencil to fall from an outstretched arm).

The 'real world' versus 'ideal world' is rather a pessimistic and simplistic view, though. Although we can never get the exact answer (to an infinite precision), we CAN get better and better approximations as our model (mathematical description) gets better and better. The problem is, the maths gets harder too, which means this is rarely developed beyond the 'constant acceleration' at A-level, or 'air resistance model' at degree level.

If you think about it, you have to get the model infinitely complicated to COMPLETELY describe every influence and factor of the real physical world. Even if you account for air resistance, you could argue some of the energy goes into rotation, so you add maths to deal with that. Then you can say that the gravity is stronger at the bottom than the top because it starts further from the centre of the Earth. Effect is tiny, obviously, but would eventually become observable as the measurements became more and more precise. Then the friction heats the air, and of course air density is a function of temperature. Convection currents. Relativistic descriptions of gravity. Deformation of the falling object itself. Earth's rotation. Pull of the sun and moon and other planets. Dark matter. Dark energy. And a million other things...

Of course, it's time to stop making the model more complicated when the agreement with experiment is sufficiently good within the precision of measurement available: in this case, within a second or so...

My only point is, the world is only non-ideal because we choose to limit the complexity of our description of it to something sensible!

Edited By The Wright Stuff on 19/05/2017 15:07:45

The Wright Stuff19/05/2017 15:27:39
avatar
908 forum posts
198 photos

This is an online calculator that is quite interesting to play with.

Putting in a (guessed) camera mass of 0.4 kg and just using the default skydiving air-resistance at the very least shows that the 7 second fall could be entirely consistent with the OP's estimate of 100 feet...

Edited By The Wright Stuff on 19/05/2017 15:27:58

TIM Shaw19/05/2017 15:28:12
162 forum posts
41 photos

Sorry TWR - wrote it this morning, then got distracted by a lunch related issue - indeed, we seem to be of the same mind, and I am a great believer in the law of diminishing returns...

My background is in Chemical Manufacturing, and I remember a particular Technical Director who was one for complicating everything to the nth degree - so much so his underlings used to say he was not a guy to let the facts get in the way of a good theory.

onetenor19/05/2017 16:16:45
1191 forum posts

So what are we going with finally 100 ft >

Mike Etheridge 119/05/2017 16:39:02
1256 forum posts
312 photos

Nice Hornets nest you have stirred up here Pete !

I will be the first to admit that the linear velocity equations may not be wholly adequate for this calculation however they give approximate results. If the plane was at 100 feet and travelling at 27 MPH, then the pilot or bottle would not fall vertically and hence the distance travelled S would exceed 100 feet and hence the angle of the fall would have to be known to determine the actual distance travelled.

Having read some of the comments I think it's relevant to add that a sprinter can cover 100 metres at about 10 seconds. Our pilot travelling at 27 MPH and accelerating downwards from 100 feet up a third of the sprinter's run must hit the deck a dam site quicker than 10 seconds !

brokenenglish19/05/2017 17:08:30
avatar
211 forum posts
21 photos

Mike, I didn't want to get involved, there's so much nonsense in all this, and you don't have the data necessary to compute an accurate answer.

However, your "horizontal velocity" consideration is totally wrong.

You're confusing "trajectory" with the vertical distance covered. When computing a time to fall from a height, only the vertical component of any motion or force is relevant. The fact that there may be horizontal deceleration has no effect at all on the vertical velocity.

Stuart Coyle19/05/2017 20:47:29
102 forum posts
4 photos

I had a quick look at this yesterday. The camera weighs 120gm.( according to Maplins). Arrived at a TV of 74m/s, so produced the second order differential equations of motion - took one look and decided to go to bed!

Mike Etheridge 119/05/2017 21:53:29
1256 forum posts
312 photos

OK BE, I will accept any criticism when I make statements about anything especially Physics and Applied Maths that i have not studied for over 50 years. Why we learnt such things that we probably would never use in our careers is a mystery and just when the chance comes you forget the important issues and hence the doubt creeps in and you make statements that are so easily criticised.

So I will never use my vague knowledge of semI- cubical hyperbolas, and La Plas Transforms etc , and if only Barnes Wallis was still alive he could fill us in on the right calculations for dropping bombs from our model aircraft. However I gather most bombs missed their targets in WW2 !

Good night SC, I think you made the right move!

.

PETER BRUCE - Eastchurch Gap19/05/2017 23:46:49
avatar
698 forum posts
104 photos

Got home near midnight so hence the late reply to wright stuff...

I will try and get the weight of the camera tomorrow and post it here. I looked at the manufacturers site but it does not give its weight so I will have to try and sort it as best I can tomorrow Regards Peter

PETER BRUCE - Eastchurch Gap20/05/2017 00:11:18
avatar
698 forum posts
104 photos

Managed to find out the weight and details of the camera....

Oregon Scientific ATC Chameleon Dual Lens Video Camera: Amazon.co.uk: Camera & Photo. ... Product Dimensions: 11.5 x 4.8 x 4 cm ; 150 g.

Hope this will help as the way some calculations are going I am thinking I need me eyes tested.

Also need to mention that the video was done in real time so I gestimate 7 seconds about right.

This has been such an interesting post as it truly combines physical properties and observations with true theoretical phisics and maths. The approximation of height = ? Will be most interesting to read the results and see the altitude ball park figure and of course how many agree.

Tom Sharp 220/05/2017 01:46:19
avatar
2367 forum posts
12 photos

Was it Archemedies what got killed by a Tortoise hitting him on the head after it was dropped by an Eagle?

KiwiKid20/05/2017 05:04:18
avatar
257 forum posts
184 photos

Just wondering - would a lipo fall faster than a camera? laugh

lipo away.jpg

All Topics | Latest Posts

Please login to post a reply.

Magazine Locator

Want the latest issue of RCM&E? Use our magazine locator link to find your nearest stockist!

Find RCM&E! 

Email News - Join our newsletter

Love Model Aircraft? Sign up to our emails for the latest news and special offers!

Latest Forum Posts
Support Our Partners
Expo Tools 14 July
Gliders Distribution
TJD Models
SHREK
Overlander
CML
Advertise With Us
Latest "For Sale" Ads
How many metres do you have to walk with your models and gear to the flying area?
Q: How many metres do you have to walk with your models and gear to the flying area?

 Less than 20m
 20 to 100m
 100 to 200m
 200 to 500m
 More than 500m

Latest Reviews
Digital Back Issues

RCM&E Digital Back Issues

Contact us

Contact us