Brian Whatcott wrote:
On Fri, 28 Jul 2006 16:03:55 -0700, Bob S
wrote:
derbyrm wrote:
With a GPS and a spring scale, you could get a pretty good approximation of
the thrust. Attach a drag; e.g. a triangle of plywood, to a line and
measure both the reduction of speed and the pounds of drag. Assume as a
first approximation that the thrust/drag curve is about linear for a small
increment of drag. (We know it's exponential, but you're probably not up
near hull speed where it goes vertical.) If there's any wind or current,
make runs in several directions.
Add drag in increments and you'll generate a series of slopes. Find an
exponential curve that fits and you should have a pretty good guess at the
thrust curve.
Roger (or maybe not)
http://home.insightbb.com/~derbyrm
"Bob S" wrote
I am not sure this will work satisfactorily. The drag total drag force
is comprised of that from the plate and that from the boat itself. I
could measure the force on the plate as you suggest with a spring scale,
but unless it is significantly greater than the drag on the boat, I
can't solve for the thrust. And while it may be much greater than the
drag on the boat as its size increases and the speed decreases, that
will be at low speed, so I still don't have thrust at full speed.
BS
Work with him, on this one. You are going [let us say] at 3.44 mph
via a gps speed average of ten runs, and then you drag a plate which
drags at 4.23 lb for an average of ten readings, which reduces your
gps speed to 3.21 mph average of ten readings. Because drag and thrust
are in balance (else you speed up or slow down) you can say it takes
4.23 lb of thrust to contribute the extra 0.23 mph from 3.21 mph
Although this one set of data is not enough, in principle you could
say
T (thrust) proportional to 3.21 ^2 kT = 3.21 ^2
and also, T + 4.23 proportional to 3.44 ^2 k(T + 4.23) = 3.44 ^2
Supposing that the scaling constant k stays constant over this small
range, we can divide one equation by t'other to get this:
(T + 4.23) / T = 3.44 ^2 / 3.21 ^2 = 1.148
T + 4.23 = 1.148T
0.148T = 4.23
T = 28.6 lb. at 3.21 mph
Like that, but with more data points.
Brian Whatcott Altus OK
Thanks for your input, Brian. Unfortunately k is not constant. The
problem is that
T1 = [k-boat + k-plate]*v1^2
If we eliminate the plate
T2 = T1 + del-T = [k-boat]*v2^2
Dividing, we get
T1/[T1 + del-T] = {[k-boat + k-plate]/k-boat}*{v1/v2}^2
I can measure del-T as you suggest as I can the velocities, and solve
for T1. But I can not accurately predict the individual drag
coefficients or their ratio. One might assume k-plate is much, much less
than k-boat (resulting in your equation), or assume it is some
particular per cent of k-boat, but then I am not sure that the result
would be any better than assuming the factory-rated thrust is realistic.
And, of course, we also have to face up to the inaccuracies of the speed
and thrust measurements.
Conversely, one could use two different sized plates, each of which had
much, much greater drag than the boat, assume the drag differs only by
the ratio of the plate areas, and solve. But again, I am afraid this is
no more accurate than the factory spec. To get the plate drag much
greater than the boat drag, you would also be in a very low speed regime
where the static thrust might be as good as whatever one measures.
Static thrust could be measured fairly accurately with a spring scale by
pivoting the motor on its mounts from a stationary point(a dock)and
measuring the force needed to hold the motor vertical.
BS