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

You're getting away from what I suggested. My idea was that the drag of the
plate was much much less than that of the boat. Measuring at various speeds
got the slope of the curve, dV/dD at those speeds. (Velocity and Drag)
It's a backward, graphic approach to integral calculus and you have the
advantage of knowing the type of curve.

Also, if your voltmeter is reasonably precise, you do have an ammeter. Just
record the voltage drop across the cables feeding the motor. Again, you
don't have nice text book values unless you put a known load on at some
point, but you can get the percentage change in current. (The voltage drop
between no-load and your load is proportional to current and to the
relatively constant resistance of the battery, connections, and cables.
Ohm's law you know.)

Roger

http://home.insightbb.com/~derbyrm

"Bob S" wrote in message
...
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

derbyrm wrote:
You're getting away from what I suggested. My idea was that the drag of the
plate was much much less than that of the boat. Measuring at various speeds
got the slope of the curve, dV/dD at those speeds. (Velocity and Drag)
It's a backward, graphic approach to integral calculus and you have the
advantage of knowing the type of curve.

Also, if your voltmeter is reasonably precise, you do have an ammeter. Just
record the voltage drop across the cables feeding the motor. Again, you
don't have nice text book values unless you put a known load on at some
point, but you can get the percentage change in current. (The voltage drop
between no-load and your load is proportional to current and to the
relatively constant resistance of the battery, connections, and cables.
Ohm's law you know.)

Roger

http://home.insightbb.com/~derbyrm

"Bob S" wrote in message
...
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


I have to leave town for a couple of days, but I'd like to follow up on
this.

First, I think Brian and I both made a mistake in confusing thrust and
drag. In our approaches we actually allowed the thrust to change by the
amount of the drag on the plate. The correct assumption, it now seems to
me, is that thrust remains approximately constant for small changes in
drag (say, 5 lb of added plate drag when the total thrust is 50 lb).

Then T = K*v1^2
and T = (K + dK)*v2^2

The result is dK/K = (V1/v2)^2 - 1

and we can determine the ratio of the drags, but not the thrust. This
could be, however, a useful tool for determining hull drag if we know
the drag on the plate. How accurately can one predict drag on a fixed
size and shape plate?

Getting back to your idea: I can see that you can make a series of
pair-wise measurements over a small range of drags that give the
approximate slope of the velocity-drag curve at each measurement point.
Graphically connecting these straight line sections would approximate a
small portion of the velocity-drag curve, but I do not see how one gets
thrust from it without knowing the actual drag.

BS

If one is not accelerating (or decelerating), then thrust equals drag.

Roger

http://home.insightbb.com/~derbyrm

"Bob S" wrote in message
...
derbyrm wrote:
You're getting away from what I suggested. My idea was that the drag of
the plate was much much less than that of the boat. Measuring at various
speeds got the slope of the curve, dV/dD at those speeds. (Velocity and
Drag) It's a backward, graphic approach to integral calculus and you have
the advantage of knowing the type of curve.

Also, if your voltmeter is reasonably precise, you do have an ammeter.
Just record the voltage drop across the cables feeding the motor. Again,
you don't have nice text book values unless you put a known load on at
some point, but you can get the percentage change in current. (The
voltage drop between no-load and your load is proportional to current and
to the relatively constant resistance of the battery, connections, and
cables. Ohm's law you know.)

Roger

http://home.insightbb.com/~derbyrm

"Bob S" wrote in message
...
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


I have to leave town for a couple of days, but I'd like to follow up on
this.

First, I think Brian and I both made a mistake in confusing thrust and
drag. In our approaches we actually allowed the thrust to change by the
amount of the drag on the plate. The correct assumption, it now seems to
me, is that thrust remains approximately constant for small changes in
drag (say, 5 lb of added plate drag when the total thrust is 50 lb).

Then T = K*v1^2
and T = (K + dK)*v2^2

The result is dK/K = (V1/v2)^2 - 1

and we can determine the ratio of the drags, but not the thrust. This
could be, however, a useful tool for determining hull drag if we know the
drag on the plate. How accurately can one predict drag on a fixed size and
shape plate?

Getting back to your idea: I can see that you can make a series of
pair-wise measurements over a small range of drags that give the
approximate slope of the velocity-drag curve at each measurement point.
Graphically connecting these straight line sections would approximate a
small portion of the velocity-drag curve, but I do not see how one gets
thrust from it without knowing the actual drag.

BS

derbyrm wrote:
If one is not accelerating (or decelerating), then thrust equals drag.

Roger

http://home.insightbb.com/~derbyrm

Yes, of course, but both Brian and I allowed thrust to increase by the
amount of the added plate drag. That is not correct.

BS

"Bob S" wrote in message
...
derbyrm wrote:
You're getting away from what I suggested. My idea was that the drag of
the plate was much much less than that of the boat. Measuring at various
speeds got the slope of the curve, dV/dD at those speeds. (Velocity and
Drag) It's a backward, graphic approach to integral calculus and you have
the advantage of knowing the type of curve.

Also, if your voltmeter is reasonably precise, you do have an ammeter.
Just record the voltage drop across the cables feeding the motor. Again,
you don't have nice text book values unless you put a known load on at
some point, but you can get the percentage change in current. (The
voltage drop between no-load and your load is proportional to current and
to the relatively constant resistance of the battery, connections, and
cables. Ohm's law you know.)

Roger

http://home.insightbb.com/~derbyrm

"Bob S" wrote in message
...
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

I have to leave town for a couple of days, but I'd like to follow up on
this.

First, I think Brian and I both made a mistake in confusing thrust and
drag. In our approaches we actually allowed the thrust to change by the
amount of the drag on the plate. The correct assumption, it now seems to
me, is that thrust remains approximately constant for small changes in
drag (say, 5 lb of added plate drag when the total thrust is 50 lb).

Then T = K*v1^2
and T = (K + dK)*v2^2

The result is dK/K = (V1/v2)^2 - 1

and we can determine the ratio of the drags, but not the thrust. This
could be, however, a useful tool for determining hull drag if we know the
drag on the plate. How accurately can one predict drag on a fixed size and
shape plate?

Getting back to your idea: I can see that you can make a series of
pair-wise measurements over a small range of drags that give the
approximate slope of the velocity-drag curve at each measurement point.
Graphically connecting these straight line sections would approximate a
small portion of the velocity-drag curve, but I do not see how one gets
thrust from it without knowing the actual drag.

BS

Hmmm? (I'm to lazy to dig into the math.) With only small changes in drag
and the throttle held constant, it would seem to me that the thrust of the
motor would be constant. Since the thrust balances the total drag one gets
the dV/dD which should be the same as dV/dT, (Velocity, Drag, Thrust) but
I assumed that dT was zero. I do see the dilemma. Did somebody divide by
zero here?

Okay. If thrust is constant for the small change in drag, then total drag
is constant, and we're getting the equation of speed versus hull drag; i.e.
the reduction in speed correlates to reduced drag by the hull. That works.
(I think.)

Roger

http://home.insightbb.com/~derbyrm

"Bob S" wrote in message
...
derbyrm wrote:
If one is not accelerating (or decelerating), then thrust equals drag.

Roger

http://home.insightbb.com/~derbyrm

Yes, of course, but both Brian and I allowed thrust to increase by the
amount of the added plate drag. That is not correct.

BS

"Bob S" wrote in message
...
derbyrm wrote:
You're getting away from what I suggested. My idea was that the drag
of the plate was much much less than that of the boat. Measuring at
various speeds got the slope of the curve, dV/dD at those speeds.
(Velocity and Drag) It's a backward, graphic approach to integral
calculus and you have the advantage of knowing the type of curve.

Also, if your voltmeter is reasonably precise, you do have an ammeter.
Just record the voltage drop across the cables feeding the motor.
Again, you don't have nice text book values unless you put a known load
on at some point, but you can get the percentage change in current.
(The voltage drop between no-load and your load is proportional to
current and to the relatively constant resistance of the battery,
connections, and cables. Ohm's law you know.)

Roger

http://home.insightbb.com/~derbyrm

"Bob S" wrote in message
...
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

I have to leave town for a couple of days, but I'd like to follow up on
this.

First, I think Brian and I both made a mistake in confusing thrust and
drag. In our approaches we actually allowed the thrust to change by the
amount of the drag on the plate. The correct assumption, it now seems to
me, is that thrust remains approximately constant for small changes in
drag (say, 5 lb of added plate drag when the total thrust is 50 lb).

Then T = K*v1^2
and T = (K + dK)*v2^2

The result is dK/K = (V1/v2)^2 - 1

and we can determine the ratio of the drags, but not the thrust. This
could be, however, a useful tool for determining hull drag if we know
the drag on the plate. How accurately can one predict drag on a fixed
size and shape plate?

Getting back to your idea: I can see that you can make a series of
pair-wise measurements over a small range of drags that give the
approximate slope of the velocity-drag curve at each measurement point.
Graphically connecting these straight line sections would approximate a
small portion of the velocity-drag curve, but I do not see how one gets
thrust from it without knowing the actual drag.

BS