Suppose the winch is spinning at 2500 rpm. The circumference of a 3" drum is 9.42 in, so the winch will pull in 2500 * 9.42 inches each minute. That means that the line would be moving at 22.3 mph along the ground. Of course, the drum diameter and/or the rpm are probably something different, but this sounds approximately right -- you see on tv that a dog can almost catch a fake rabbit pulled by a winch.
Why then does the plane go so fast? Because its speed is a multiple of the line's ground speed, depending on the climb angle (the plane's airspeed is the ground speed of the line divided by the cosine of the launch angle). The steeper the climb, the faster the plane will be traveling. If the plane is climbing at 75 degrees, it's traveling at 86 mph.
As the plane gets to the top and starts diving, the dive speed will be increased by the acceleration of gravity and the stored energy in the stretched line.
You can make a rough guess at the plane's speed when it enters the zoom, as well as the amount of altitude gained in a zoom, by timing the duration of the zoom -- if it's vertical (if not, this doesn't work). At the WSM, I timed many zooms. For the moment, let's make a lot of simplifying assumptions. A vertical zoom that lasts 4 seconds will give up 4 * 32 ft/sec, or 128 ft/sec in velocity. If the ending velocity is 30 ft/sec, the plane had to start out at 158 ft/sec, or 108 mph. The average velocity is (30 + 158) / 2 or 94 ft/sec, so the plane gained 4 * 94 = 376 feet in the zoom.
Why then does the plane go so fast? Because its speed is a multiple of the line's ground speed, depending on the climb angle (the plane's airspeed is the ground speed of the line divided by the cosine of the launch angle). The steeper the climb, the faster the plane will be traveling. If the plane is climbing at 75 degrees, it's traveling at 86 mph.
As the plane gets to the top and starts diving, the dive speed will be increased by the acceleration of gravity and the stored energy in the stretched line.
You can make a rough guess at the plane's speed when it enters the zoom, as well as the amount of altitude gained in a zoom, by timing the duration of the zoom -- if it's vertical (if not, this doesn't work). At the WSM, I timed many zooms. For the moment, let's make a lot of simplifying assumptions. A vertical zoom that lasts 4 seconds will give up 4 * 32 ft/sec, or 128 ft/sec in velocity. If the ending velocity is 30 ft/sec, the plane had to start out at 158 ft/sec, or 108 mph. The average velocity is (30 + 158) / 2 or 94 ft/sec, so the plane gained 4 * 94 = 376 feet in the zoom.
----- Original Message ----
From: RBurnoski <[EMAIL PROTECTED]>
To: soaring@airage.com
Sent: Tuesday, September 26, 2006 10:03:25 AM
Subject: [RCSE] Top end speed in MPH ! Math quiz
From: RBurnoski <[EMAIL PROTECTED]>
To: soaring@airage.com
Sent: Tuesday, September 26, 2006 10:03:25 AM
Subject: [RCSE] Top end speed in MPH ! Math quiz
Please read all the factors below before
responding
For you math guys.
Lets take the NATS/ LSF winches as an
example.
OK, first , line
diameter may not be needed for this calculation but drum diameter would
be . Im not sure of that exact drum diameter,,,,,,,
???? 3" ?
If using a three inch diameter , lets keep it
constant. I know the drum gets bigger when the line is
stacked. But, you can figure it both ways if your
having a slow day.
The question is:
What is the maximum miles per hour that the
winch will go pulling a 4 lb to a 6 lb model ?
One other factor Im giving you. Im not
talking about launching using the normal arc up over the turnaround.
Im talking about a parrallel to the ground, from winch toward the turnaround
scenerio.
Thanks in
advanced,
Richard Burnoski
