i havent priced a 30lb, but off the top of my head, a no name made in china
13lb trolling motor is about AU$200 while a 50lb is about AU$500. add in
battery, wiring etc and i could buy a brand new petrol outboard for those
prices. I think the electric outboards are still really expensive in
australia for what they are. they will probably come down in price in a
year or two but for now i'd feel a bit cheated paying for one.

Shaun


Should also have mentioned that my 30 lb Endura pushes a 14' Hobie at
about 3.5 mph in still water.

BS

Shaun Van Poecke wrote:
i havent priced a 30lb, but off the top of my head, a no name made in china
13lb trolling motor is about AU$200 while a 50lb is about AU$500. add in
battery, wiring etc and i could buy a brand new petrol outboard for those
prices. I think the electric outboards are still really expensive in
australia for what they are. they will probably come down in price in a
year or two but for now i'd feel a bit cheated paying for one.

Shaun

Should also have mentioned that my 30 lb Endura pushes a 14' Hobie at
about 3.5 mph in still water.

BS


That does seem rather high. My 30 lb was $99 and the 50 Lb $130. But
they work really well, appear to be well made, and are very reliable.

BS

Should also have mentioned that my 30 lb Endura pushes a 14' Hobie at
about 3.5 mph in still water.

I think this general theme (Using inboard pumps for propulsion) is
worth pursuing. One interesting aspect is being able to use the same
pump(s) for forward propulsion and bow-thruster/steering.

Can we start with a decent idea of the efficiency of the trolling
motors?

How much current do some typical units draw (all 12Volts?) ??

Are they rated only in static thrust? Or also Horsepower? Horsepower
can be converted to Force VS Distance VS Time. (1.0 Horsepower == 550
Foot-Pounds per second, right??) HighSchool Physics was, um, 50 years
ago :-) Yes, I just see "1 horsepower [electric] = 550.221 382 975
foot pound-force/second" at
http://www.onlineconversion.com/power.htm

So, IF you knew the relationship of Drag (In Pounds) VS Speed for YOUR
boat, you could create a graph of Horsepower VS Speed. (This would
be for "Perfect Horsepower" which certainly will not happen with real-
word trolling motors and propellers, OR real-world pumps and
hoses.. )

But you'd have SOME idea...

If you had a 25 pound fish scale and 100 feet of line, and someone to
paddle the boat OUT so you could pull it IN, you might start to get
some numbers...

Other Related Idea: I have thought about running a medium-large (??)
pump from my inboard boat engine to bow ports for "Bow Thruster".
Anyone seen something like this?? A pump could be engaged with an Air
Conditioner Clutch....

So let's keep thinking about this????

wrote:
Should also have mentioned that my 30 lb Endura pushes a 14' Hobie at
about 3.5 mph in still water.

I think this general theme (Using inboard pumps for propulsion) is
worth pursuing. One interesting aspect is being able to use the same
pump(s) for forward propulsion and bow-thruster/steering.

Can we start with a decent idea of the efficiency of the trolling
motors?

How much current do some typical units draw (all 12Volts?) ??

Are they rated only in static thrust? Or also Horsepower? Horsepower
can be converted to Force VS Distance VS Time. (1.0 Horsepower == 550
Foot-Pounds per second, right??) HighSchool Physics was, um, 50 years
ago :-) Yes, I just see "1 horsepower [electric] = 550.221 382 975
foot pound-force/second" at
http://www.onlineconversion.com/power.htm

So, IF you knew the relationship of Drag (In Pounds) VS Speed for YOUR
boat, you could create a graph of Horsepower VS Speed. (This would
be for "Perfect Horsepower" which certainly will not happen with real-
word trolling motors and propellers, OR real-world pumps and
hoses.. )

But you'd have SOME idea...

If you had a 25 pound fish scale and 100 feet of line, and someone to
paddle the boat OUT so you could pull it IN, you might start to get
some numbers...

Other Related Idea: I have thought about running a medium-large (??)
pump from my inboard boat engine to bow ports for "Bow Thruster".
Anyone seen something like this?? A pump could be engaged with an Air
Conditioner Clutch....

So let's keep thinking about this????


Generally speaking a figure of about 50% overall efficiency (prop +
motor) seems to be appropriate for my 30 and 50 lb motors if you believe
the factory-stated thrust and current, based on the speeds I've
measured. Current draw is 30 and 42 amps respectively. There are a few
(very expensive) motors that do better. Motors are available the run on
12, 24, 36 and 48 volts. Generally trolling motors are rated in static
thrust while "electric outboards" are quoted in hp. It is difficult to
directly compare electric hp with gas because the gas motor hp differs
tremendously with rpm. That having been said, volts x amps is still true
input power, and 550 ft-lb/sec is hp regardless of how it is generated.

BS

R Swarts wrote:
wrote:

Should also have mentioned that my 30 lb Endura pushes a 14' Hobie at
about 3.5 mph in still water.


I think this general theme (Using inboard pumps for propulsion) is
worth pursuing. One interesting aspect is being able to use the same
pump(s) for forward propulsion and bow-thruster/steering.


I would tend to doubt that, but...

Can we start with a decent idea of the efficiency of the trolling
motors?

How much current do some typical units draw (all 12Volts?) ??

Are they rated only in static thrust? Or also Horsepower? Horsepower
can be converted to Force VS Distance VS Time. (1.0 Horsepower == 550
Foot-Pounds per second, right??) HighSchool Physics was, um, 50 years
ago :-) Yes, I just see "1 horsepower [electric] = 550.221 382 975
foot pound-force/second" at
http://www.onlineconversion.com/power.htm

Well, yes, but HP is really irrelevant other than as a comparison
between similarly configured devices. What you're really concerned with
is *Force*, calculated as F=ma (i.e. force = mass x acceleration). No
matter what system you employ for propulsion, it boils down to the mass
of water displaced per unit time.

To do a meaningful comparison, you'd need to know a lot of information.
You need to have a force chart for the outboard (i.e. mass flow rate
generated by the propeller over the operational range, versus amp draw
for the motor). Then you could compare the amp draw of your pump versus
mass flow rate. I think you'll find that the pump idea is *far* less
effcient than an outboard. The propeller has no frictional losses
associated with supplying water to, or discharging water from, the
'pumping' device. The frictional loss at the propeller surface is
offset by the surface frictional losses at the pump impeller.

Keith Hughes

On Feb 15, 5:47 pm, Keith Hughes wrote:
R Swarts wrote:

Well, yes, but HP is really irrelevant other than as a comparison
between similarly configured devices. What you're really concerned with
is *Force*, calculated as F=ma (i.e. force = mass x acceleration). No
matter what system you employ for propulsion, it boils down to the mass
of water displaced per unit time.


Keith, I think you're missing a factor here. "Displaced" implies a
DISTANCE per unit time. That's where horsepower (Or any other POWER
unit) matters. 550 FOOT - (Pounds-Force) (Per SECOND) means that you
could "Gear Down" (or use other force multiplication arrangements) and
lift 550 pounds at 1 foot per second, or instead lift 55 pounds at 10
feet per second. Right?

A large diameter 4-blade prop on a 60 foot boat with a small pitch and
a 20 HP diesel with a reduction gear can put a (lot) of FORCE on that
boat and move it at 5 or 6 knots. I've seen an old 20 Hp Mercury
outboard push a 3-point Hydro at over 40 MPH.

We don't know enough here (yet) to say what the efficiency of an
inboard pump would be in moving a small boat against it's frictional
resistance at a certain speed. We all know, from experience, that
moving a small boat like the Hobie mentioned at the beginning of this
thread takes VERY little force at very small speeds. A gentle push by
hand moves it right away from the dock. Most of us have moved a 10
meter or larger boat a few feet by leaning a little on a dockline.

What we're missing is some approximation of the efficiency of a well-
designed pump in converting electrical power to mechanical power to
move a boat. I think we'd have to do some research and talk to some
mechanical engineers who understand pumps! I have a friend who
recently built a small Hydroelectric plant in New York, who did his
own calculations and is using a large (Thing formerly sold as a pump)
as a turbine, with excellent efficiency. He's running 2 typical homes
on it.. It's a BIG pump and he's at the bottom of a 85 foot
waterfall...

Someone somewhere knows a lot more about propelling a boat with an
inboard pump than I do!

Ive been reading up a bit on pumps, but some of the math is beyond me. i do
know that its possilb eo hook must pumps up either in series, or in
parallel. in parallel you quite logically get a doubling of flow in gallons
per hour or whatever, while in series you combine the 'heads' whatever that
means. i think it means head pressure?

i know a lot of the losses in small pumps are from pumping 'up'. most small
pumps are rated by how high they can pump water, and the rating for flow
goes down as the height increases. installed in a boat, i would try to keep
the whole thing on the level with the shortest hose runs possible. on a
beach cat, i would have a thru hull on the side of the hull with maybe 6
inches of hose going to the pump, then another foot of hose going to the
outlet.

i think youd have to start with two pumps in each hull, both running off a
common larger diameter inlest, and through a Y joiner to a common outlet.
this would give you some options. you could run the pumps in parallel, or
in series. then you would have to experiement with various reductions in
the outlet to see what the smallest diameter nozzle you could use without
losing flow would be. this is probably how you would use 'gearing'.

if you used too large of a diameter nozzle, you really wouldnt get any force
at all.

Shaun



Keith, I think you're missing a factor here. "Displaced" implies a
DISTANCE per unit time. That's where horsepower (Or any other POWER
unit) matters. 550 FOOT - (Pounds-Force) (Per SECOND) means that you
could "Gear Down" (or use other force multiplication arrangements) and
lift 550 pounds at 1 foot per second, or instead lift 55 pounds at 10
feet per second. Right?

A large diameter 4-blade prop on a 60 foot boat with a small pitch and
a 20 HP diesel with a reduction gear can put a (lot) of FORCE on that
boat and move it at 5 or 6 knots. I've seen an old 20 Hp Mercury
outboard push a 3-point Hydro at over 40 MPH.

We don't know enough here (yet) to say what the efficiency of an
inboard pump would be in moving a small boat against it's frictional
resistance at a certain speed. We all know, from experience, that
moving a small boat like the Hobie mentioned at the beginning of this
thread takes VERY little force at very small speeds. A gentle push by
hand moves it right away from the dock. Most of us have moved a 10
meter or larger boat a few feet by leaning a little on a dockline.

What we're missing is some approximation of the efficiency of a well-
designed pump in converting electrical power to mechanical power to
move a boat. I think we'd have to do some research and talk to some
mechanical engineers who understand pumps! I have a friend who
recently built a small Hydroelectric plant in New York, who did his
own calculations and is using a large (Thing formerly sold as a pump)
as a turbine, with excellent efficiency. He's running 2 typical homes
on it.. It's a BIG pump and he's at the bottom of a 85 foot
waterfall...

Someone somewhere knows a lot more about propelling a boat with an
inboard pump than I do!

wrote:
On Feb 15, 5:47 pm, Keith Hughes wrote:

R Swarts wrote:


Well, yes, but HP is really irrelevant other than as a comparison
between similarly configured devices. What you're really concerned with
is *Force*, calculated as F=ma (i.e. force = mass x acceleration). No
matter what system you employ for propulsion, it boils down to the mass
of water displaced per unit time.


Keith, I think you're missing a factor here. "Displaced" implies a
DISTANCE per unit time. That's where horsepower (Or any other POWER
unit) matters. 550 FOOT - (Pounds-Force) (Per SECOND) means that you
could "Gear Down" (or use other force multiplication arrangements) and
lift 550 pounds at 1 foot per second, or instead lift 55 pounds at 10
feet per second. Right?

I'm not saying that HP is not a valid unit of measure, just that
comparing HP for two different propulsion types, with what could be
vastly different efficiencies, is of little practical value.
"Displaced" was probably not the clearest of terms to use. I was not
referring to 'volume' displaced by the boat through the water, rather
the 'mass' of water accelerated (assuming a fixed exit velocity), per
unit time, by the propulsion device (prop or what have you).

A large diameter 4-blade prop on a 60 foot boat with a small pitch and
a 20 HP diesel with a reduction gear can put a (lot) of FORCE on that
boat and move it at 5 or 6 knots. I've seen an old 20 Hp Mercury
outboard push a 3-point Hydro at over 40 MPH.

In both cases, however, the only FORCE available comes from accelerating
a mass of water.


We don't know enough here (yet) to say what the efficiency of an
inboard pump would be in moving a small boat against it's frictional
resistance at a certain speed.

The boats' frictional resistance is a fixed quantity, and therefore
irrelevant for comparison of propulsion types.

We all know, from experience, that
moving a small boat like the Hobie mentioned at the beginning of this
thread takes VERY little force at very small speeds. A gentle push by
hand moves it right away from the dock. Most of us have moved a 10
meter or larger boat a few feet by leaning a little on a dockline.

What we're missing is some approximation of the efficiency of a well-
designed pump in converting electrical power to mechanical power to
move a boat.

However efficient the 'pump' is, it still has the added frictional
losses associated with the system (suction line, pump casing, discharge
line) that the outboard prop does not. Also, keep in mind that a Bilge
pump is *not* designed for this application, so it's a safe bet that the
resulting efficiency will be sub-optimal at best. OTOH, the outboard
*is* designed for the application.

Clearly this type of pumping system *can* work for propulsion, and for
any given pump you just have to look at the pumping curve (for the
suction and discharge head, which wouldn't be too difficult to
calculate). From that, you can calculate the resulting mass flowrate,
and from that the force applied. The bigger problem I see is
characterizing the "flow", if you will (mass of water accelerated, and
the acceleration applied), of the outboard prop versus energy input. A
curve which will almost certainly vary significantly with boat speed.

I think we'd have to do some research and talk to some
mechanical engineers who understand pumps! I have a friend who
recently built a small Hydroelectric plant in New York, who did his
own calculations and is using a large (Thing formerly sold as a pump)
as a turbine, with excellent efficiency.

Excellent "results" don't necessarily imply efficiency.

He's running 2 typical homes
on it.. It's a BIG pump and he's at the bottom of a 85 foot
waterfall...

Next time I find myself with an 85' waterfall in my backyard, I'll have
to give it a try. 'Course if you have a spare one laying around that you
want to part with cheap... ;-)

Keith Hughes

Hi Shaun,

Ive been reading up a bit on pumps, but some of the math is beyond me. i do
know that its possilb eo hook must pumps up either in series, or in
parallel. in parallel you quite logically get a doubling of flow in gallons
per hour or whatever, while in series you combine the 'heads' whatever that
means. i think it means head pressure?

Yep, head is pressure. Basically, you have one pound/sq.inch for each
27.68" of water column (height).

i know a lot of the losses in small pumps are from pumping 'up'.

That's kind of a misconception regarding 'head'. Pumping up, down, or
horizontal, the flowrate is dependent on the total backpressure on the
discharge line (but of course, 10' of vertical pipe does have more total
backpressure than 10' of horizontal pipe - of the same size).

most small
pumps are rated by how high they can pump water, and the rating for flow
goes down as the height increases. installed in a boat, i would try to keep
the whole thing on the level with the shortest hose runs possible. on a
beach cat, i would have a thru hull on the side of the hull with maybe 6
inches of hose going to the pump, then another foot of hose going to the
outlet.

Keeping the tubing runs as short as possible is certainly the right
approach to reduce frictional losses. One problem with the inlet on the
side of the hull (or any hull surface tangential to the water flow) is
that you get Bernoulli effects as the boat speed increases, that tends
to create a vacuum in the suction line (the same concept that makes
paint sprayers - the kind that use air hoses - or end-of-hose garden
sprayers work. The high speed stream across the diptube end creates
suction to raise the paint/roundup into the discharge stream).

i think youd have to start with two pumps in each hull, both running off a
common larger diameter inlest, and through a Y joiner to a common outlet.
this would give you some options. you could run the pumps in parallel, or
in series. then you would have to experiement with various reductions in
the outlet to see what the smallest diameter nozzle you could use without
losing flow would be. this is probably how you would use 'gearing'.

if you used too large of a diameter nozzle, you really wouldnt get any force
at all.

Don't confuse "velocity" with "Force". Just like with a garden hose
where you have, say 80psig, you can pinch the end to get a higher
velocity stream, but you get less flow (i.e. less mass). Since the
force = mass x acceleration, the force however is the same (you only
have 80psig to start with). The same is true for pumps, as you note
above, when you create more backpressure (pinching the hose), the
flowrate goes down. If you move 100gpm of water through the system, the
force is the same whether the discharge is 1" or 3", only the velocity
of the dischage changes. Remember, PSI is pounds per square inch (i.e.
force per unit area), so the 1" discharge stream may be at 10 times the
pressure of the 3" stream, but the 3" stream has 10 times the
cross-sectional area of the 1" stream.


Keith Hughes

The following was posted by Cliff on the psubs.org group. Personal
submarine and ROV builders consider bilge pumps frequently. They are
often used in ROV's because they are easy, but bilge pumps or jet
propulsion is rarely used in Subs because they are inefficient. If
you do go with a pump, be sure to place the discharge just above the
water line to increase the efficiently. --Doug www.submarineboat.com

Below is a derivation of thrust that can be developed from a axial
flow pump
in terms of volumetric flowrate.
The thrust due to accelerating fluid through a pump can be written as
F=M(V1-V0)
Where M is the mass flow rate, V0 is the free stream velocity upstream
of
the pump and V1 is the velocity exiting the pump.
But the mass flow rate M can be related to the volumetric flow rate Q
as
M=Density*Q
Substituting, the thrust in terms of volumetric flow rate is
F=Density*Q(V1-V0)
But the volumetric flow rate Q is related to velocity in the pump duct
ID as
Q=V1*A=V1*Pi*D^2/4
Where D is the duct ID.
Solving for V1, and substituting, the thrust can be written as
F=Density*Q(Q/(Pi*D^2)-V0)
For a thruster oriented approximately normal to the direction of flow,
the
inlet velocity can be assumed to be zero. The thrust then reduces to
F = 4*Density*Q^2/(Pi*D^2)
Or
F= 0.001766*(q/d)^2
for freshwater where,
F = Thrust, lbf
q = pump volumetric flow rate in gpm
d = pump outlet duct inside diameter in inches
As an example, a pump with a capacity of 200 gpm flowing through a 2"
duct
would develop 17.7 lbf of thrust.