Electric outboards
 

I have noticed that real numbers concerning electric outboards seem to
be few and far between. Therefore, I thought some of you might be
interested in some results I obtained the other day with mine.

I used two Excide 6 volt golf cart batteries in series. Open circuit
voltage at the time of the tests was 12.53 volts.

The motors were both Minnkota Enduras, one 30 lb, the other 50 lb. I
have no way of directly measuring thrust, nor did I have a calibrated
current shunt, so I assumed the factory-published values of 30 lb at 30
amps and 50 lb at 42 amps.

Voltages at the input to the motor leads were 12.14 and 11.96 with the
motors set to max. This indicates and combined internal battery and
external wiring resistance of about .013 ohm. The nominal input powers
are therefore .49 hp and .67 hp.

Again, assuming factory stated thrust is accurate, the output powers are
..28 hp and .53 hp for overall efficiencies of 57% for the 30 pounder and
80% for the 50. Incidentally, when I questioned Minnkota by phone they
would not state efficiencies but did say the 50 lb unit is their most
efficient.

These motors push a 16 ft flat-bottomed sailing skiff at 3.5 and 4.0 mph
respectively.

I would be pleased to hear from any of you who are also interested in
electric propulsion.

Bob Swarts

Electric outboards
 

On Wed, 26 Jul 2006 08:43:23 -0700 Bob S
) wrote:
I have noticed that real numbers concerning electric outboards seem to
be few and far between. Therefore, I thought some of you might be
interested in some results I obtained the other day with mine.

I used two Excide 6 volt golf cart batteries in series. Open circuit
voltage at the time of the tests was 12.53 volts.

The motors were both Minnkota Enduras, one 30 lb, the other 50 lb. I
have no way of directly measuring thrust, nor did I have a calibrated
current shunt, so I assumed the factory-published values of 30 lb at 30
amps and 50 lb at 42 amps.

Voltages at the input to the motor leads were 12.14 and 11.96 with the
motors set to max. This indicates and combined internal battery and
external wiring resistance of about .013 ohm. The nominal input powers
are therefore .49 hp and .67 hp.

Again, assuming factory stated thrust is accurate, the output powers are
.28 hp and .53 hp for overall efficiencies of 57% for the 30 pounder and
80% for the 50. Incidentally, when I questioned Minnkota by phone they
would not state efficiencies but did say the 50 lb unit is their most
efficient.

These motors push a 16 ft flat-bottomed sailing skiff at 3.5 and 4.0 mph
respectively.

See the extra power needed for that 0.5 mile faster?

I would be pleased to hear from any of you who are also interested in
electric propulsion.

The Queen Mary II has electric propulsion.;-) For now I think battery
propulsed boats will be a niche product, nice for fishing or cruising
protected areas or coming home but not for holidays.

Do the math and find out yourself. My boat has also electric propulsion.
When the day comes that fuelcell or other electric generation becomes
cheaper then a diesel generator set, I'm prepared. Untill then the
diesel keeps running.

--
Richard
e-mail: vervang/replace invalid door/with NL.net
http://web.inter.nl.net/users/schnecke/

Electric outboards
 

Richard van den Berg wrote:
On Wed, 26 Jul 2006 08:43:23 -0700 Bob S
) wrote:
I have noticed that real numbers concerning electric outboards seem to
be few and far between. Therefore, I thought some of you might be
interested in some results I obtained the other day with mine.

I used two Excide 6 volt golf cart batteries in series. Open circuit
voltage at the time of the tests was 12.53 volts.

The motors were both Minnkota Enduras, one 30 lb, the other 50 lb. I
have no way of directly measuring thrust, nor did I have a calibrated
current shunt, so I assumed the factory-published values of 30 lb at 30
amps and 50 lb at 42 amps.

Voltages at the input to the motor leads were 12.14 and 11.96 with the
motors set to max. This indicates and combined internal battery and
external wiring resistance of about .013 ohm. The nominal input powers
are therefore .49 hp and .67 hp.

Again, assuming factory stated thrust is accurate, the output powers are
.28 hp and .53 hp for overall efficiencies of 57% for the 30 pounder and
80% for the 50. Incidentally, when I questioned Minnkota by phone they
would not state efficiencies but did say the 50 lb unit is their most
efficient.

These motors push a 16 ft flat-bottomed sailing skiff at 3.5 and 4.0 mph
respectively.

See the extra power needed for that 0.5 mile faster?

I would be pleased to hear from any of you who are also interested in
electric propulsion.

The Queen Mary II has electric propulsion.;-) For now I think battery
propulsed boats will be a niche product, nice for fishing or cruising
protected areas or coming home but not for holidays.

Do the math and find out yourself. My boat has also electric propulsion.
When the day comes that fuelcell or other electric generation becomes
cheaper then a diesel generator set, I'm prepared. Untill then the
diesel keeps running.

Bear in mind that many of the boats on the water today ARE used for
fishing or utilitarian functions on protected waters. No one suggested
serious cruising. But my 16-footer does have a 30 mile range at 4 mph at
the 70% discharge level which is more than adequate for its crabbing and
homecoming (sail backup) functions. The math does work in many cases,
but, as you say, picking the right niche is important.

BS

Electric outboards
 

On Wed, 26 Jul 2006 08:43:23 -0700, Bob S
wrote:

I have noticed that real numbers concerning electric outboards seem to
be few and far between. Therefore, I thought some of you might be
interested in some results I obtained the other day with mine.

I used two Excide 6 volt golf cart batteries in series. Open circuit
voltage at the time of the tests was 12.53 volts.

The motors were both Minnkota Enduras, one 30 lb, the other 50 lb. I
have no way of directly measuring thrust, nor did I have a calibrated
current shunt, so I assumed the factory-published values of 30 lb at 30
amps and 50 lb at 42 amps.

Voltages at the input to the motor leads were 12.14 and 11.96 with the
motors set to max. This indicates and combined internal battery and
external wiring resistance of about .013 ohm. The nominal input powers
are therefore .49 hp and .67 hp.

Again, assuming factory stated thrust is accurate, the output powers are
.28 hp and .53 hp for overall efficiencies of 57% for the 30 pounder and
80% for the 50. Incidentally, when I questioned Minnkota by phone they
would not state efficiencies but did say the 50 lb unit is their most
efficient.

These motors push a 16 ft flat-bottomed sailing skiff at 3.5 and 4.0 mph
respectively.

I would be pleased to hear from any of you who are also interested in
electric propulsion.

Bob Swarts


Small electric motors can have 80% efficiency. Small water propellers
can have 80% efficiency.
If you take the product of voltage across the motor, AT the motor, and
the current through the motor, and multiply by 0.64 and divide by 746
you'll have an estimate of the net HP available for thrust.

V x I x 0.64 / 746 = HP for thrust.
Thrust at constant power varies with water speed, and is greatest at
standstill
(which is why troll motor makers specify thrust at standstill, where
it is meaningless)

Let's work your numbers:
12.14V x 30A x 0.64 / 746 = 0.3 HP
11.96V x 42A x 0.64 / 746 = 0.43 HP

Rough, rough cross check:
if power required is proportional to v^3
then power required at 3.5 mph is 3.5 x 3.5 x 3.5 / ( 4 x 4 x 4) or
0.67 of power at 4 mph.
Power available for thrust at 3.5 mph = 0.3/0.43 = 0.88 of power at 4
mph

This suggests to me (it could be a dozen other things) that the prop
is less optimal on the faster skiff.


Take it with a pinch


Brian Whatcott Altus OK

Electric outboards
 

Brian Whatcott wrote:
On Wed, 26 Jul 2006 08:43:23 -0700, Bob S
wrote:

I have noticed that real numbers concerning electric outboards seem to
be few and far between. Therefore, I thought some of you might be
interested in some results I obtained the other day with mine.

I used two Excide 6 volt golf cart batteries in series. Open circuit
voltage at the time of the tests was 12.53 volts.

The motors were both Minnkota Enduras, one 30 lb, the other 50 lb. I
have no way of directly measuring thrust, nor did I have a calibrated
current shunt, so I assumed the factory-published values of 30 lb at 30
amps and 50 lb at 42 amps.

Voltages at the input to the motor leads were 12.14 and 11.96 with the
motors set to max. This indicates and combined internal battery and
external wiring resistance of about .013 ohm. The nominal input powers
are therefore .49 hp and .67 hp.

Again, assuming factory stated thrust is accurate, the output powers are
.28 hp and .53 hp for overall efficiencies of 57% for the 30 pounder and
80% for the 50. Incidentally, when I questioned Minnkota by phone they
would not state efficiencies but did say the 50 lb unit is their most
efficient.

These motors push a 16 ft flat-bottomed sailing skiff at 3.5 and 4.0 mph
respectively.

I would be pleased to hear from any of you who are also interested in
electric propulsion.

Bob Swarts


Small electric motors can have 80% efficiency. Small water propellers
can have 80% efficiency.
If you take the product of voltage across the motor, AT the motor, and
the current through the motor, and multiply by 0.64 and divide by 746
you'll have an estimate of the net HP available for thrust.

V x I x 0.64 / 746 = HP for thrust.
Thrust at constant power varies with water speed, and is greatest at
standstill
(which is why troll motor makers specify thrust at standstill, where
it is meaningless)

Let's work your numbers:
12.14V x 30A x 0.64 / 746 = 0.3 HP
11.96V x 42A x 0.64 / 746 = 0.43 HP

Rough, rough cross check:
if power required is proportional to v^3
then power required at 3.5 mph is 3.5 x 3.5 x 3.5 / ( 4 x 4 x 4) or
0.67 of power at 4 mph.
Power available for thrust at 3.5 mph = 0.3/0.43 = 0.88 of power at 4
mph

This suggests to me (it could be a dozen other things) that the prop
is less optimal on the faster skiff.


Take it with a pinch


Brian Whatcott Altus OK

Basically I agree with your analysis, but you assume equal over all
efficiencies for both motors, which negates both Minnkota's published
specs and my observations. Modern small electric motors can (not saying
this is actually the case)have efficiencies well above 90%. One of the
problems is that neither Minnkota nor MotorGuide will state efficiencies
for their motors or props. But IF this is the case with the Minnkota 50
lb model, then an over all efficiency of 80% would not be impossible. In
any case, without being able to accurately measure the current or thrust
at speed, it is difficult to wring much more information from my
results. Also, there were probably a couple hundredths of a volt more
drop along the internal and external motor cable (I measured at the
input end of the factory cable). Note, though, that your theoretical
values of available hp (.3 and .43) are not in bad agreement with my
observed results of .28 and .53 hp.

I sure would like to see some more people contribute some actual numbers.

BS

Electric outboards
 

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 in message
...
Brian Whatcott wrote:
On Wed, 26 Jul 2006 08:43:23 -0700, Bob S
wrote:

I have noticed that real numbers concerning electric outboards seem to
be few and far between. Therefore, I thought some of you might be
interested in some results I obtained the other day with mine.

I used two Excide 6 volt golf cart batteries in series. Open circuit
voltage at the time of the tests was 12.53 volts.

The motors were both Minnkota Enduras, one 30 lb, the other 50 lb. I
have no way of directly measuring thrust, nor did I have a calibrated
current shunt, so I assumed the factory-published values of 30 lb at 30
amps and 50 lb at 42 amps.

Voltages at the input to the motor leads were 12.14 and 11.96 with the
motors set to max. This indicates and combined internal battery and
external wiring resistance of about .013 ohm. The nominal input powers
are therefore .49 hp and .67 hp.

Again, assuming factory stated thrust is accurate, the output powers are
.28 hp and .53 hp for overall efficiencies of 57% for the 30 pounder and
80% for the 50. Incidentally, when I questioned Minnkota by phone they
would not state efficiencies but did say the 50 lb unit is their most
efficient.

These motors push a 16 ft flat-bottomed sailing skiff at 3.5 and 4.0 mph
respectively.

I would be pleased to hear from any of you who are also interested in
electric propulsion.

Bob Swarts


Small electric motors can have 80% efficiency. Small water propellers
can have 80% efficiency.
If you take the product of voltage across the motor, AT the motor, and
the current through the motor, and multiply by 0.64 and divide by 746
you'll have an estimate of the net HP available for thrust.

V x I x 0.64 / 746 = HP for thrust.
Thrust at constant power varies with water speed, and is greatest at
standstill
(which is why troll motor makers specify thrust at standstill, where
it is meaningless)

Let's work your numbers:
12.14V x 30A x 0.64 / 746 = 0.3 HP
11.96V x 42A x 0.64 / 746 = 0.43 HP

Rough, rough cross check:
if power required is proportional to v^3
then power required at 3.5 mph is 3.5 x 3.5 x 3.5 / ( 4 x 4 x 4) or
0.67 of power at 4 mph.
Power available for thrust at 3.5 mph = 0.3/0.43 = 0.88 of power at 4
mph

This suggests to me (it could be a dozen other things) that the prop
is less optimal on the faster skiff.


Take it with a pinch


Brian Whatcott Altus OK

Basically I agree with your analysis, but you assume equal over all
efficiencies for both motors, which negates both Minnkota's published
specs and my observations. Modern small electric motors can (not saying
this is actually the case)have efficiencies well above 90%. One of the
problems is that neither Minnkota nor MotorGuide will state efficiencies
for their motors or props. But IF this is the case with the Minnkota 50 lb
model, then an over all efficiency of 80% would not be impossible. In any
case, without being able to accurately measure the current or thrust at
speed, it is difficult to wring much more information from my results.
Also, there were probably a couple hundredths of a volt more drop along
the internal and external motor cable (I measured at the input end of the
factory cable). Note, though, that your theoretical values of available hp
(.3 and .43) are not in bad agreement with my observed results of .28 and
.53 hp.

I sure would like to see some more people contribute some actual numbers.

BS

Electric outboards
 

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 in message
...
Brian Whatcott wrote:
On Wed, 26 Jul 2006 08:43:23 -0700, Bob S
wrote:

I have noticed that real numbers concerning electric outboards seem to
be few and far between. Therefore, I thought some of you might be
interested in some results I obtained the other day with mine.

I used two Excide 6 volt golf cart batteries in series. Open circuit
voltage at the time of the tests was 12.53 volts.

The motors were both Minnkota Enduras, one 30 lb, the other 50 lb. I
have no way of directly measuring thrust, nor did I have a calibrated
current shunt, so I assumed the factory-published values of 30 lb at 30
amps and 50 lb at 42 amps.

Voltages at the input to the motor leads were 12.14 and 11.96 with the
motors set to max. This indicates and combined internal battery and
external wiring resistance of about .013 ohm. The nominal input powers
are therefore .49 hp and .67 hp.

Again, assuming factory stated thrust is accurate, the output powers are
.28 hp and .53 hp for overall efficiencies of 57% for the 30 pounder and
80% for the 50. Incidentally, when I questioned Minnkota by phone they
would not state efficiencies but did say the 50 lb unit is their most
efficient.

These motors push a 16 ft flat-bottomed sailing skiff at 3.5 and 4.0 mph
respectively.

I would be pleased to hear from any of you who are also interested in
electric propulsion.

Bob Swarts

Small electric motors can have 80% efficiency. Small water propellers
can have 80% efficiency.
If you take the product of voltage across the motor, AT the motor, and
the current through the motor, and multiply by 0.64 and divide by 746
you'll have an estimate of the net HP available for thrust.

V x I x 0.64 / 746 = HP for thrust.
Thrust at constant power varies with water speed, and is greatest at
standstill
(which is why troll motor makers specify thrust at standstill, where
it is meaningless)

Let's work your numbers:
12.14V x 30A x 0.64 / 746 = 0.3 HP
11.96V x 42A x 0.64 / 746 = 0.43 HP

Rough, rough cross check:
if power required is proportional to v^3
then power required at 3.5 mph is 3.5 x 3.5 x 3.5 / ( 4 x 4 x 4) or
0.67 of power at 4 mph.
Power available for thrust at 3.5 mph = 0.3/0.43 = 0.88 of power at 4
mph

This suggests to me (it could be a dozen other things) that the prop
is less optimal on the faster skiff.


Take it with a pinch


Brian Whatcott Altus OK

Basically I agree with your analysis, but you assume equal over all
efficiencies for both motors, which negates both Minnkota's published
specs and my observations. Modern small electric motors can (not saying
this is actually the case)have efficiencies well above 90%. One of the
problems is that neither Minnkota nor MotorGuide will state efficiencies
for their motors or props. But IF this is the case with the Minnkota 50 lb
model, then an over all efficiency of 80% would not be impossible. In any
case, without being able to accurately measure the current or thrust at
speed, it is difficult to wring much more information from my results.
Also, there were probably a couple hundredths of a volt more drop along
the internal and external motor cable (I measured at the input end of the
factory cable). Note, though, that your theoretical values of available hp
(.3 and .43) are not in bad agreement with my observed results of .28 and
.53 hp.

I sure would like to see some more people contribute some actual numbers.

BS


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

Electric outboards
 

Nice post! Just as practical as it could be.

Brian Whatcott Altus OK

On Fri, 28 Jul 2006 12:46:44 -0400, "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 in message
...
Brian Whatcott wrote:
On Wed, 26 Jul 2006 08:43:23 -0700, Bob S
wrote:

I have noticed that real numbers concerning electric outboards seem to
be few and far between. Therefore, I thought some of you might be
interested in some results I obtained the other day with mine.

I used two Excide 6 volt golf cart batteries in series. Open circuit
voltage at the time of the tests was 12.53 volts.

The motors were both Minnkota Enduras, one 30 lb, the other 50 lb. I
have no way of directly measuring thrust, nor did I have a calibrated
current shunt, so I assumed the factory-published values of 30 lb at 30
amps and 50 lb at 42 amps.

Voltages at the input to the motor leads were 12.14 and 11.96 with the
motors set to max. This indicates and combined internal battery and
external wiring resistance of about .013 ohm. The nominal input powers
are therefore .49 hp and .67 hp.

Again, assuming factory stated thrust is accurate, the output powers are
.28 hp and .53 hp for overall efficiencies of 57% for the 30 pounder and
80% for the 50. Incidentally, when I questioned Minnkota by phone they
would not state efficiencies but did say the 50 lb unit is their most
efficient.

These motors push a 16 ft flat-bottomed sailing skiff at 3.5 and 4.0 mph
respectively.

I would be pleased to hear from any of you who are also interested in
electric propulsion.

Bob Swarts


Small electric motors can have 80% efficiency. Small water propellers
can have 80% efficiency.
If you take the product of voltage across the motor, AT the motor, and
the current through the motor, and multiply by 0.64 and divide by 746
you'll have an estimate of the net HP available for thrust.

V x I x 0.64 / 746 = HP for thrust.
Thrust at constant power varies with water speed, and is greatest at
standstill
(which is why troll motor makers specify thrust at standstill, where
it is meaningless)

Let's work your numbers:
12.14V x 30A x 0.64 / 746 = 0.3 HP
11.96V x 42A x 0.64 / 746 = 0.43 HP

Rough, rough cross check:
if power required is proportional to v^3
then power required at 3.5 mph is 3.5 x 3.5 x 3.5 / ( 4 x 4 x 4) or
0.67 of power at 4 mph.
Power available for thrust at 3.5 mph = 0.3/0.43 = 0.88 of power at 4
mph

This suggests to me (it could be a dozen other things) that the prop
is less optimal on the faster skiff.


Take it with a pinch


Brian Whatcott Altus OK

Basically I agree with your analysis, but you assume equal over all
efficiencies for both motors, which negates both Minnkota's published
specs and my observations. Modern small electric motors can (not saying
this is actually the case)have efficiencies well above 90%. One of the
problems is that neither Minnkota nor MotorGuide will state efficiencies
for their motors or props. But IF this is the case with the Minnkota 50 lb
model, then an over all efficiency of 80% would not be impossible. In any
case, without being able to accurately measure the current or thrust at
speed, it is difficult to wring much more information from my results.
Also, there were probably a couple hundredths of a volt more drop along
the internal and external motor cable (I measured at the input end of the
factory cable). Note, though, that your theoretical values of available hp
(.3 and .43) are not in bad agreement with my observed results of .28 and
.53 hp.

I sure would like to see some more people contribute some actual numbers.

BS

Electric outboards
 

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

Electric outboards
 

On Fri, 28 Jul 2006 02:36:00 GMT Brian Whatcott
) wrote:

Small electric motors can have 80% efficiency. Small water propellers
can have 80% efficiency.
If you take the product of voltage across the motor, AT the motor, and
the current through the motor, and multiply by 0.64 and divide by 746
you'll have an estimate of the net HP available for thrust.

V x I x 0.64 / 746 = HP for thrust.
Thrust at constant power varies with water speed, and is greatest at
standstill
(which is why troll motor makers specify thrust at standstill, where
it is meaningless)

Thrust depends on the propeller dimension. Bollard pull (Google gives
plenty hits) is commonly used with tugs, but conditions have to meet
minimal requirements to measure.

My previous boat was (minimal) faster with its 11 hp then a identical
boat with 16 hp. Bollard pull with my boat give a 150 kg, the boat with
16 hp 100 kg.

Let's work your numbers:
12.14V x 30A x 0.64 / 746 = 0.3 HP
11.96V x 42A x 0.64 / 746 = 0.43 HP

Rough, rough cross check:
if power required is proportional to v^3
then power required at 3.5 mph is 3.5 x 3.5 x 3.5 / ( 4 x 4 x 4) or
0.67 of power at 4 mph.
Power available for thrust at 3.5 mph = 0.3/0.43 = 0.88 of power at 4
mph

Thrust and boat speed don't behave proportional, why else bollard pull
for tugs.

This suggests to me (it could be a dozen other things) that the prop
is less optimal on the faster skiff.

Possible indeed. I think efficiency is best tested by measuring the
supplied power at different speeds, but requires an ammeter.

--
Richard
e-mail: vervang/replace invalid door/with NL.net
http://web.inter.nl.net/users/schnecke/

Electric outboards
 

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

Electric outboards
 

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

Electric outboards
 

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

Electric outboards
 

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

Electric outboards
 

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

Electric outboards
 

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