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10-10-2005 11:51 AM  12 years agoPost 1
trvo

rrVeteran

Bognor Regis,UK, Aurorra Ltd

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Hey guys,

I am just bored at the mo, last weekend my raptor 90 fell 250ft with a stopped rotor (690 V's) (blade stop went wrong). It weighs 4.5Kg, how can I calculate roughly how fast it was falling when it hit the ground?

Cheers - Trev.

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Trex 600e
Trex 600n
Aurorra Ltd, Align UK, Align Factory, CSM, Quick UK, Heli-Lessons

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10-10-2005 11:54 AM  12 years agoPost 2
nivlek

rrProfessor

Norfolk England

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How many times you were able to swear while the heli fell multiplied by
the depth of the crater in cm + half you age = MPH .

At the end of the day , it gets dark .

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10-10-2005 12:17 PM  12 years agoPost 3
rchelifreak

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Melbourne, Australia --M.R.C.H.C. and K.D.M.A.S.--

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im running around in circles cos i should know this crap

got my physics exam in like 2 weeks
well i got the force that it hit the ground with is 44.1N and thats about as far as i got

hahahahahah
ill try do it later



Don't drink and drive. You might hit a bump and spill your drink.
Adrian
:D

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10-10-2005 12:21 PM  12 years agoPost 4
tris_heli

rrKey Veteran

Lidlington, Bedfordshire, UK

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Doesn’t it depend on the "drag" or air resistance of the model ?? (thinking that if it had a parachute handing out, it would have loads more drag and probably be in one piece today)...


But the more important question, where's the pictures ????

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10-10-2005 12:22 PM  12 years agoPost 5
trvo

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Bognor Regis,UK, Aurorra Ltd

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Nivlek, using your formulae I hit the ground at 349.5 MPH.

68*5+9.5 = 349.5

Trev

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Trex 600n
Aurorra Ltd, Align UK, Align Factory, CSM, Quick UK, Heli-Lessons

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10-10-2005 12:24 PM  12 years agoPost 6
trvo

rrVeteran

Bognor Regis,UK, Aurorra Ltd

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Pics are here:

Trev

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Trex 600e
Trex 600n
Aurorra Ltd, Align UK, Align Factory, CSM, Quick UK, Heli-Lessons

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10-10-2005 12:26 PM  12 years agoPost 7
rchelifreak

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Melbourne, Australia --M.R.C.H.C. and K.D.M.A.S.--

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ignoring air resistence, i think i got it:
x=83.3 m
u=0
v=?
a=-9.8ms

v= 40.406ms
=145.46 km/h

i think

maths ppl jump me if im wrong plz



Don't drink and drive. You might hit a bump and spill your drink.
Adrian
:D

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10-10-2005 12:28 PM  12 years agoPost 8
trvo

rrVeteran

Bognor Regis,UK, Aurorra Ltd

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cool, rchelifreak, a fairly reasonable speed then!

Trev

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Trex 600e
Trex 600n
Aurorra Ltd, Align UK, Align Factory, CSM, Quick UK, Heli-Lessons

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10-10-2005 12:31 PM  12 years agoPost 9
rchelifreak

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Melbourne, Australia --M.R.C.H.C. and K.D.M.A.S.--

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i think thats right

thats ignoring air resistence/drag but im pretty sure its close

LMAO
nice photo



Don't drink and drive. You might hit a bump and spill your drink.
Adrian
:D

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10-10-2005 01:10 PM  12 years agoPost 10
Ed Moore

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West Sussex, UK

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That'll be much reduced with air resistance- on a random note did you know that when cycling above 20mph, on flat ground, 90% of the resistance is air resistance, i.e. 90% of your energy is expended by overcoming drag. Scary.

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10-10-2005 01:11 PM  12 years agoPost 11
ChrisMoore

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Bay Village, OH

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You can use the law of conservation of energy to find out how fast it was going when it hit (ignoring air resistance, of course). That would let us find that the Initial kenetic energy plus the potential energy equals the final kinetic energy. setting up an equation like this :

mgh+(1/2)mv.^2=(1/2)mv'^2

In this instance I think that the helicopter was not at a dead stop at 250 feet, I imagine it had some velocity pointed earth-wards (toward earth will be negative and away would be postive so the value of the initial velocity would have to follow those parameters for calcuating).

I have represented the initial velocity as "v." and the final velocity as " v' " We can plug in values for everything else we know (m=4.5kg; g=-9.8m/s^2 and h= 250feet =76.2 meters) Everything has to be metric! in our case, mass cancels out, so we dont even 'need' to know 4.5kg.

so we can solve for v' and get:

v'= sqrt(2gh+v.^2)

If the initial velocity was zero, which I dont know for sure then we could find the final velocity by setting v.=0 :

v'=sqrt(2mgh)=sqrt(2(-9.8m/s^2)(76.2m))=38.65m/s which is pretty quick (86.45 mph or 139.14 kmh).


Thats my shot at it, I think I have everything.


Chris

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10-10-2005 05:02 PM  12 years agoPost 12
melsman

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Atascadero, CA

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That's all of the damage!? Hardly a scratch! Little more than heavy hangar-rash.

Looks like it's ready to go for another blade-stop auto. That is, if you can throw it high enough...

Ashley

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10-10-2005 08:14 PM  12 years agoPost 13
rplumbo

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St. Paul, MN

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Here's the right way to do this...what you're looking for is the terminal velocity of your R90. Meaning, your heli is no longer accelerating in its fall, but it's falling as fast as drag will allow it to.

Here's the math:
In FFF (Fast Free Fall) the SUM of the forces on the heli is ZERO. F=ma, and the acceleration is zero because it's falling at a constant velocity. Remember? It's falling as fast as it possibly can.

So...what's pulling on the heli? Gravity. What's pushing on the heli? Drag. If the sum of the forces is zero, then they must equal each other.
Then, D = mg

Well, what's D? We know m (mass of the heli in kg), and we know g (9.81 m/s/s)

D = q*S*Cd
Where:
q = 1/2*density of air * velocity of the fall squared
S = area of heli presented to the air flow, so assuming no rotation, it's the area parallel with the ground.
Cd = Coefficient of Drag. This is a measured parameter from windtunnel tests or other mathematical build up methods, but using .2-.4 is a good approximation.

So plugging all this in to the above, we have...
1/2*(density of air)*(velocity)^2*(Wetted area)*(Coefficient of Drag) = m*g

Solve this for velocity and you get:
V = (2*m*g/(density of air * Cd * Wetted Area))^1/2

So, try it...density of air is 1.19 kg/m^3 at sea level. The area of a falling heli is probably something like .2 or .3 meters^2, Cd of .2...see what you get.

I get like 35 m/s which is 78 mph. This will depend on your choices for area, Cd and mass. Play with it and you'll get a feel for the ball park.

Reid

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10-10-2005 08:18 PM  12 years agoPost 14
nivlek

rrProfessor

Norfolk England

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I've got a radar gun . If you'd like to come over to my flying site , you can do it again and I'll try to clock it for ya !

At the end of the day , it gets dark .

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10-11-2005 05:24 AM  12 years agoPost 15
Vitya

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North York, Ontario

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78 mph is much more reasonable. The speeds ignoring drag are not usefull as it makes a big difference, I mean, imagine skydiving with no drag...

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