A Few Words About VRS

[Wally]

 

In more technical terms we’re referring to the ideal hover induced velocity (v_h) which is: v_h(ft/sec) = √(T/2ρA). Were T= thrust, ρ= is air density, and A= rotor disk area.

 

However we can break that down to make the math easier:

v_{h(ft/sec) =} (√1/2ρ) x (√T/A)

 

To reduce it even more, we’ll reference our work to sea level standard day.

 

Air density at sea level is .002377 slugs per cubic foot. How we can solve the first part (√1/2ρ) = √(1/.004754) = √(210.34) = 14.5

 

We’re left with 14.5√T/A. Since thrust must equal weight in any steady state flight condition, we can substitute the gross weight for the term “T”

 

Now all that remains is 14.5√(gross wt./disk area) or simply 14.5√(DL)

 

Just makes the math easier.

 

I get a lesser induced velocity out of this as altitude increases... can't be right?

[iChris]

 

I get a lesser induced velocity out of this as altitude increases... can't be right?

 

OK, lets go back to the math. I think you’re looking at the 14.5 in the following relationship:

 

14.5 = ρ = air density (sea level), so if air density decreases the number 14.5 must decrease and that means less induced velocity.

 

It doesn’t work that way. You can’t increase and decrease your coefficient that way. You must go back a step. Remember how we derived the number 14.5, which is our coefficient in math terms.

 

In math, both sides of the equal sign (=) must maintain their proportional relationship, so we need to recalculate the coefficient.

 

If we go back to how we derived 14.5 we see the following:

 

√1/2ρ = 14.5 Consequently, we have an inverse ratio, in other words, ρ is inversely proportional to the induced flow and the coefficient (14.5). Therefore, as the air density decreases the induced flow and the coefficient must increase.

 

Sea level standard day (ρ=. 002377):

√ (1/2ρ) = √ (1/. 004754) = 14.5

 

10,000’ standard day (ρ=. 001755)

√ (1/2ρ) = √ (1/. 00351) = 16.87

 

That’s the increase in coefficient you were looking for.

[Wally]

My bad assumption. Is this it, paraphrased?

Induced velocity (fps) = square root of (weight/2*(slugs)*disk area)

 

ISA zero alt

"x"= square root(4961/(2*0.00237717*306.75), so X = 58.32 fps, 3499 fpm

 

ISA 12000 (already had the "ρ" in calculator)

"x"= square root(4961/(2*0.00164779*306.75), so X = 70.05 fps, 4203 fpm

 

That intuits as approximation, back and forth...

Edited February 21, 2014 by Wally

[aeroscout]

Not that I'm complaining, but this few words about VRS has grown a bit. It never hurts too much for me to learn more though.

[iChris]

My bad assumption. Is this it, paraphrased?

Induced velocity (fps) = square root of (weight/2*(slugs)*disk area)

 

ISA zero alt

"x"= square root(4961/(2*0.00237717*306.75), so X = 58.32 fps, 3499 fpm

 

ISA 12000 (already had the "ρ" in calculator)

"x"= square root(4961/(2*0.00164779*306.75), so X = 70.05 fps, 4203 fpm

 

That intuits as approximation, back and forth...

 

Correct, that’s it…

 

As a side note, since we were talking about VRS we referenced the induced velocity at the rotor disc (v₁ or v_h); however, the max velocity (v₂) is in the remote wake below the rotor that reaches twice (2X) the air velocity at the rotor disc.

 

Stagnant air is being taken from above the rotor and accelerated down through the rotor disk to its final velocity below the rotor, the momentum theory.

[pilot#476398]

"A few words about VRS"?

 

Load Disk Early

[aeroscout]

"A few words about VRS"?

 

Load Disk Early

A man of few (power packed) words !

[Nearly Retired]

Forget the hard numbers. Forget 'em! I mean, they're great and all...and you *should* know them...I guess...

 

But...

 

What if you're making a nice steep approach to a helipad on the side of a hill...not at the top but somewhere down from the peak. You're coming in, below ETL now, but all fat, dumb and happy...knowing that you're keeping your rate-of-descent just below the Forbidden Threshhold...

 

...And you suddenly experience an updraft. Or a downdraft.

 

Now the actual rate of descent of the fuselage means not a damn thing. You are not in control of the air flowing through your rotor. Because nature is...obviously...chaotic and you never know what the wind is going to do.

 

So yeah. Knowing the numbers is great. Memorize iChris's theories if you like. But it might not matter a whit out in the real world.

[WolftalonID]

Non of that math had +'s or -'s and so it looked like greek to me anywhooo