REFLECTOR: Florida shares its wealth

Fri Mar 11 07:08:01 CST 2005

Greg,

A couple more hurricanes (ne: cyclones) and you'll have to check your map to make sure your not in Florida.

Chuck

-----Original Message-----
From: reflector-bounces at tvbf.org [mailto:reflector-bounces at tvbf.org]On Behalf Of Greg Poole
Sent: Thursday, March 10, 2005 11:28 PM
To: Velocity Aircraft Owners and Builders list
Subject: Fw: REFLECTOR: Re: Falling objects through the prop (Was NOSEWHEELGOTSIDEWAYS)

Oh....and sorry for such a technical explanation!

G!

----- Original Message ----- 
From: Greg Poole <mailto:gpoole at zeta.org.au>  
To: Velocity Aircraft Owners and Builders list <mailto:reflector at tvbf.org>  
Sent: Friday, March 11, 2005 3:26 PM
Subject: Re: REFLECTOR: Re: Falling objects through the prop (Was NOSE WHEELGOTSIDEWAYS)

Maybe it did and it got lucky & was not effectively "strained"?

;^)

Greg in Sydney

----- Original Message ----- 
From: Tom Martino <mailto:tmartino at troubleshooter.com>  
To: Velocity Aircraft Owners and Builders  <mailto:reflector at tvbf.org> list 
Sent: Friday, March 11, 2005 3:09 PM
Subject: RE: REFLECTOR: Re: Falling objects through the prop (Was NOSE WHEELGOTSIDEWAYS)

Except ... it didn't.

-----Original Message-----
From: reflector-bounces at tvbf.org [mailto:reflector-bounces at tvbf.org] On Behalf Of KeithHallsten
Sent: Thursday, March 10, 2005 6:48 PM
To: Velocity Aircraft Owners and Builders list
Subject: REFLECTOR: Re: Falling objects through the prop (Was NOSE WHEEL GOTSIDEWAYS)

Al,

I follow you up through the point where you say the object will only fall 85 msec before the prop comes by.  However, as I recall the position formula is:

   X = X(init) + V(init)*T + 1/2 A*T*T.   If we use the initial elevation of the object as our datum, X(init) = 0, and if the initial downward velocity, V(init), is also zero.  If we further assume that the only acceleration acting on the object is gravity (32.2 ft/sec^2), then

  Distance Fallen = 1/2 * (32.2 ft/sec^2) * (0.085 sec) * (0.085 sec) = 0.117 ft  = 1.40 inch.

So does your prop arc extend lower than 1.4 inches below the level of your nose gear door?  Sure!  

However, we have neglected the fact that any object that falls off the plane is initially at the speed of the plane, not at rest, and it is slowed by air friction, so it is decelerating horizontally while is is accelerating vertically downward.  If we know the drag coefficient of the object that's falling, we could calculate that deceleration due to the relative wind, but that will be left as an exercise for the reader!  Suffice it to say that the object will actually fall significantly longer that 85 msec, and will therefore fall farther.  If the prop arc extends, say, 16 inches below the belly of the plane, we would need to have approximately 1.0 second of fall to clear the prop arc.  Is this likely?  Probably not.

I think the conclusion is that anything that falls off the fuselage, even from the nose, is pretty likely to pass through the prop arc.

Keith

----- Original Message ----- 

From: Al Gietzen <mailto:ALVentures at cox.net>  

To: 'Velocity Aircraft  <mailto:reflector at tvbf.org> Owners and Builders list' 

Sent: Thursday, March 10, 2005 8:08 AM

Subject: RE: REFLECTOR: RE: NOSE WHEEL GOT SIDEWAYS

Subject: REFLECTOR: RE: NOSE WHEEL GOT SIDEWAYS

Tom

You say believed you lost the "T" fitting in flight.  Did you check the Velocity high speed rotating strainer that's attached to the engine to see if it was detected, evidenced by leaving 'memory marks'?  Idle curiousity:  Can anything fall off the plane's fuselage without going through the prop at rotation speeds or above?

Chuck

Well, let's see; at 120 mph that's 176 ft/sec - so it takes about 85 msec to travel 15 ft.  If you drop something from the nose gear well, it will travel downward a max of about 30 - 35 in. before the prop goes by.  It might clear the prop.

Just thinking out loud, check my math.

Al

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