Thrust measurements

[Fastjeff57]

Dumb question: Anyone know how to calculate the thrust in pounds of force developed by an outboard motor? (Electric OBs are rated in pounds of force, but not in hp.) I'm looking at designing a tilt system and need to know what kind of force it has to resist.

Thanks,

Jeff

[Mark75H]

As I understand it, you want to calculate how much thrust the motor is applying to the water so you can know how much force is being applied to the assembly you want to build.

If you learn how much force you are making at the prop ... you will still have to make a correction at the bracket location ... half way between the prop and the tilt pivot will have twice the force being applied to the transom.

Subtracting "slip" as has been suggested to you on another forum is incorrect. Slip is force that the prop applies but is "consumed" by the boat as friction against the water - its still part of the push against the transom

[david bryan]

other factors need to be addressed i would not expect you would have as much force on the transom on 100 pound boat as you would on 1000 pound boat unless you where tied to the dock. But look a the Mercury kick outs they have done lots of research

[Fastjeff57]

Sam:

Thanks for the come back. Calculating the force on the transom is hte easy part! What I need is the force produced by the prop. I double checked Min Kota's e-motor site for kWs as well as thrust, but no luck.

Gotta be a way to do this!

Jeff

PS to David: There's no difference in thrust on a 100 or a 1000 lb boat if the motor is at full song. One just goes a lot faster.

[MN1]

Does this help?
Mark N
http://sites.mercurymarine.com/porta..._schema=PORTAL

[Seagull 170]

As I understood it, slip is the difference between the theoretical speed & the actual speed

i.e. if you have a 12" pitch prop running at 5000 revs, your theoretical speed would be 12" x 5000 x 60 = inches per hour which when turned in miles per hour = 56.8 miles per hour

But the speed you record at 5000 revs in your boat using a 12" pitch prop, on your GPS is 45mph, your slip is 20.7%, which is caused by a combination of hull drag & prop inefficiency.

& I think you will find that if your prop is exerting 100 lbs thrust into the water it will put the same thrust 100 lbs onto the transom. As the motor is not acting as a lever but as a fixed arm, otherwise we would all be using the longest leg motors you could find!

Your answer is only right Sam, if the pivot point is attached to a fixed point, not on the boat.

Perhaps Ron would be the best qualified to pass comment on this!

[zul8tr]

[QUOTE=Seagull 170;85810]As I understood it, slip is the difference between the theoretical speed & the actual speed

i.e. if you have a 12" pitch prop running at 5000 revs, your theoretical speed would be 12" x 5000 x 60 = inches per hour which when turned in miles per hour = 56.8 miles per hour

But the speed you record at 5000 revs in your boat using a 12" pitch prop, on your GPS is 45mph, your slip is 20.7%, which is caused by a combination of hull drag & prop inefficiency.

& I think you will find that if your prop is exerting 100 lbs thrust into the water it will put the same thrust 100 lbs onto the transom. As the motor is not acting as a lever but as a fixed arm, otherwise we would all be using the longest leg motors you could find!

Your answer is only right Sam, if the pivot point is attached to a fixed point, not on the boat.

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The prop thrust will not be 100% transferred to the transom because there are other forces on the lower unit when underway. When the motor and boat is on the beach the horizontal load transfered to the transom kick out bracket is just due to the weight of the motor acting at its center of gravity at a distance 'L' from the clamp pivot to the rear of the transom. There is a moment of Engine weight x L that is balanced by Transom force at kick out x vertical distance from kick out bracket to the clamp pivot. From this the horizontal force at the kick out can be easily calculated.

When the boat is underway there are dynamic forces that enter the calc. These are the thrust and its leverage arm to the clamp pivot and opposing the thrust is the drag on the gear case,skeg and the rotating prop airfoil. This drag force will reduce the effect of the thrust force that transferes to the transom kickout bracket. As speed increases toward terminal max on very light fast planing hulls the thrust load gets reduced quite a bit by these lower unit and prop drag forces. This is ONE reason that there are engine tie downs to the transom to prevent engine bounce up at the transom that will effect performance. Some old timer figured this out way back.

[david bryan]

Sam:

Thanks for the come back. Calculating the force on the transom is hte easy part! What I need is the force produced by the prop. I double checked Min Kota's e-motor site for kWs as well as thrust, but no luck.

Gotta be a way to do this!

Jeff

PS to David: There's no difference in thrust on a 100 or a 1000 lb boat if the motor is at full song. One just goes a lot faster.

funny transoms are torn out not pushed in

[Fastjeff57]

Mark nailed it!

Prop thrust = 375 x Hp (output)/ boat speed.

Sample calculation: 47 lbs of thrust for a 50 hp motor pushing a boat 40 mph. (Not much force.)

And I was wrong about the force on the transom of a 100 lb vs. a 1,000 lb boat, even at full throttle, for when boat speed goes down, force goes up.

Thanks all,

Jeff