I'll have a pop...

This is particularly easy to overcomplicate, so at the simplest and most uncomplicated level 'every action has an equal and opposite reaction'...

Say your chopper weighs 1300Kg
1m^3 of air weighs 1.3Kg
1000m^3 of air weighs 1300Kg

Gravity is trying to accelerate you downwards at 9.8ms^2
you have to accelerate 1000m^3 of air through the blades at 9.8ms^2 to counter it.

Or accelerate 2000m^3 at 4.9ms^2...etc

The pressure during hover (ideally) will be the weight of the chopper over the area of the disc.


Does that work?

No way, Way no Aro....
 

Well ...
the mass per second (M/T)

M/T = Square root (2WpA) (Crab)

so if you quadruple the weight you only double the Mass per second

Intuitively this makes sense cos if you double the mass per time(independantly) then you'll ALSO have to be making it go twice as fast -

so you don't need to double the mass per time if you double your weight !

.... a very different answer to ARO - I am simplifying it - just not over-simplifying!

[p is density 1.3kg/m^3 W is weight in Newtons ie AUMass times g, (10Newtons/kg), A area], Formulae unverified

and Vi= Square root(2W/pA) (Crab)

Aro - you converted the mass of the helo to a force in newtons eg 1300kg x 9.8m/s = 12740 newtons mass x acceleration = force - therefore IMHO you should do the same to the air eg 1.3kg x 9.8 m/s = 12.7 newtons because the air is at rest before you push it through the disc and is still affected by the gravitational constant.

crab, he did.

Hi Q,

You said earlier...
"F = Vi M/T = Vi pA Vi/2 = 1/2 pAVi^2 = W (W is weight of heli)"

You've proved Force (mKgs^-2) equals Weight(or Mass?) (Kg)

I think it should read...

Fr = Vi M/T = Vi pA Vi/2 = 1/2 pAVi^2 = Fh

Where Fr is Force required and Fh is Force imparted by the heli.

Or...The force required to hold the heli is the same as the force the heli imparts.

Pressure is newtons per metre squared.

Say the disk is 10m^2 and the heli is 1300kg, gravity is 9.8m/s^2.

F=12740N
A=10m^2

Pressure=12740/10
=1274Pa

or the Force of the chopper over the area of the disk.

This demonstrates that a smaller disk will create a higher pressure for the same chopper, due to the increased work rate, and also concurs with Vi being dependent on the radius of the disk.


There is an issue with the following equations...

No1
M/T = pAV (p is density Rho, A area of disc, and V is Vi/2 'cos that's the speed at the disc ie the 'average speed')

and

No2
M/T = Square root (2WpA) (Crab)


Lets first find Vi with our (aribtrary) numbers

vi = [(Mass x 9.81) / (2 x Density of air x Disk Area)]^0.5
vi = [(1300 x 9.8) / (2 x 1.3 x 10)]^.05
vi = (12740 / 26) ^.5
vi = 22m/s

No 1
M/T = pAV
M/T = 1.3 x 10 x 11
M/T = 286Kg/s

No 2
M/T = Square root (2WpA)
M/T = (2x1300x1.3x10)^.5
M/T = 33800^.5
M/T = 183kg/s

If we say
No 3
M/T = sqrt(2FpA)
M/T = (2x12740x1.3x10)^.5
M/T = 576Kg/s

I'm inclined to go with the last one which is an equivalent of 443m^3/s. (For a 1300kg chopper with a 10m^2 disk).

To provide a force equal to that of the chopper, that air would need to be accelerated at...


1000/443=2.25

2.25 * 9.8 = 22m/s^2

Which would see the air at vi after 1 second.


My apologies for the misuse of 'weight' in the previous post.

I wish I hadn't asked now.:confused:

Doh! My brain hurts.

I often get asked how much downwash there is in the hover.

I normally use the following formula:

(Dese5tonsachopperup = 5tonsadatairdown = lotsawindsoplizdon'tweardemhatsorputupdoseumbrellasuntildewh oledamnlotisondegroundandstopped) squared.

It normally helps for passengers and suffices on a practical basis for pilots of simple brain like wot mine is. :rolleyes:

Phuddled... worth ignoring!
 

Aro this is forensic Math !
- But your Fisiks is phukedup (not that mine isn't a bit also)


"You've proved Force (mKgs^-2) equals Weight(or Mass?) (Kg) "

No, no, no, ... yes:

Weight is a force! The OTHER expression for gravity is helpful ie. g= 9.8 N/kg
Weight isn't in Kg! Weight is Mass x g which is N (Newtons) so mKgs^-2 = N = (g x Kg)

So Fr = Fh = F = W = Mass x 9.8(N/Kg or ms^-2)

So 'your' "No 2" is "No 3" !!



Is this you trying to quote me?:
"vi = [(Mass x 9.81) / (2 x Density of air x Disk Area)]^0.5 "

if so it was:
Vi=(2W/pA)^0.5 (The 2 is on the top)

So
Vi = [(2x1300x9.8)/(1.3x10)]^0.5
= [25480/13]^.5 = 44 m/s Not 22 m/s



No 1
M/T = pAV
M/T = 1.3 x 10 x 22 (you've got 11 where I've got 22 BUT....)
M/T = 286Kg/s !!!!surprisingly same as your answer!!!
!!!you got there by making a mistake!!!


No 2
M/T = Square root (2WpA) or W=F so "No 2" is "No 3"
M/T = (2x1300x9.8x1.3x10)^.5
= 331240^.5
M/T = 575 Kg/s

But MY mistake this should have read:
M/T = Square root (0.5xWpA)

So No 2 which is the same as No 3
Gave an answer which was Square root(2/0.5) Too big so should have been:
M/T = 575/2 = 287.5 Kg/s which is the same as No 1 (rounding errors)


So No 1 = No 2 = No 3 !!!!!!


But to summaries (very clumsily) it's almost exactly what Nick Lappos elegantly said:

the 287.5 Kg/s is NOT "1300kg of air being accelerated cos thats the mass of the helicopter"
like most people wrongly assert.


Aro
so 287.5Kg/s is equivalent to 221m^3/s

Are you saying:

1000m^3, which is NOT the amount of air this helicopter is acting on per second,

is 1000/221 = 4.52 times the amount which it IS acting on per second

and so if it were acting on 221m^/s instead of 1000 m^3/s it would have to accelerate it at g x 4.52 = 9.8 x 4.52 = 44m/s^2 which happens to be the correct flow speed?

?

If so I think you are wrong (and right Post Script).

I think you are effectively calculating the factor by which you are wrong and multiplying your wrong answer by the correcting factor here.

effectively you are saying if you took a mass of air equal to the mass of the helicopter and multiplied g by the factor that that mass of air was incorrect by you'd get the correct flow speed...


Yea yea yea - I've got it! sorry I thought you were answering the question ... you're right in as far as you go (except for the basic errors ( factor of 2) which we both made.

What a waste of time - but didn't you find some of the relationships ineresting tho?

Like:

"Vi depends on 1/r ! "
and
"if you Quadruple the Area you Halve the Vi "

The easy way to look at the physics is to recall that the force is proportional to the time rate of change of the momentum (the real wording of Newton's second law). Don't use the acceleration, use the velocity change on a mass flow (a delta V on a mass/sec)

If the air has a mass M (the density times the volume) and you give it a change in velocity of V, and it is flowing through the disk at that velocity, then the force or lift generated is MV/T.

Use mass in slugs (a unit of 32.2 pounds), assume a lift of 20,000 pounds, and that air has a density of .002378 slugs/cubic foot. Assume a downwash velocity of about 35 knots (60 ft/sec)

20000=60 x M/T

The mass flow for the air = M/T = 333 slugs per second=10,000 lb/sec, which is about 140,000 cubic feet of air per second. If the column of air is 53 feet in diameter ( or 2200 square feet, the size of the disk) it would have a velocity of 63 ft/sec (which is almost exactly what we assumed).
So, to keep a 20,000 pound helo hovering, if the rotor has a 53 foot diameter, the mass that flows per second is 10,000 pounds of air, and it has a final velocity of 60 ft/sec.

In other words, if the helicopter had a disk loading of 9 pounds per square foot, it has to throw half its weight in air EVERY SECOND to stay up, and it throws that air at 60 feet per second downward.