Re Post #14: Hear hear Roy👍

Re Post #15: Hear hear Ronald👍
Doug😎

Hi Roy,

"Hi Alessandro I shall be in Australia this time next week."
You travel faster than the internet, lucky you, ahahahahahah.

Hi Roy, sorry for the digression.

I don't want to create constants.
This practical experiment (which I would have already done if I had had everything I needed) will give easily interpretable results.
In my opinion the results will be significant and not misleading.
All parameters will be the same: weight applied to the winch, supply voltage, technical data of the motor. We are not interested in reading the actual number of revolutions, we will only read the value of the electric current because from that I will understand which engine is working harder.
Which motor is demanding more current from the battery and is therefore more likely to heat up.
Logically by rpm, I mean the nameplate data, not the actual data.

Hi Alessandro I shall be in Australia this time next week.
Roy

Hi Ron, forgive me, you're right but I had this curiosity.

It's true the topic is different but it's not the first time it's explored other fields, before me there's a long list of confessed criminals who have already committed this crime, ahahahahaha.

Roy cited an A+ from his son and I couldn't resist.
It's a wonderful thing to hear how things work in other parts of the planet, not on the internet but using the internet (the internet as a medium and not as a source of information).
Is it the same in Canada as in the UK that you know?

To think that in Australia or Canada there is someone who is building boats like me and I can ask him for information directly is fantastic.

Alexandro

Your education posting doesn’t fit with …gears to match speeds… how about removing it and begin a new topic. Say “what is the educational path in your area of the planet ?”

I am a retired teacher and in our faculty of education course on the philosophy of education there are many varieties that have been developed over the years.

But, this topic doesn’t fit under the forum topic of “Using gears to match speed”

My opinion

Hi Alessandro re post 11.
I think you are creating constants that do not stand up with a load involved. If the motor drives a prop of given size then there will be a particular rpm.

With a geared motor the motor will spin faster as there is less load from the prop per revolution of the motor, so the geared motor prop speed is probably slower but unkown. Only the load produced by the prop has remained constant.
The prop tube and shaft will have a different effect due to differences in speed and therefore friction.
The identical motor is performing in a different part of its operating spectrum so although the same motor, its performance is different.

However that is why we have some rules of thumb which cover an expected average performance. The boat will be in varying conditions, open water wind small protected pools and even the depth of water. So with these unknowns we make our initial choices.

So you can still do your research and testing but the weather conditions often determine events. I play for safety with extra power available if needed.

With my yachts my test at home before they go into the water is to turn them upside down to see if all the controls inside still stay in place.

My dynamometer experiments detailed before were really of personal interest and also to measure o/p of some steam engines built by a friend.

My email system was out for 2 days due to the fighting which broke out in my laptop where 2 security systems were suddenly active. They were fighting each other and slowly destroyed all around them! It had to have professional help to restore a working system. That seems to be a metaphor for what is going on in the world at present!

Roy

Good morning everyone.

Leaving the opinion of Roy and all the modelers pending and awaiting the practical experiment that I intend to do as soon as possible, the question of A+ as a grade came to mind.

I was intrigued and I would like to know how it works in the UK and if it is also like this in Anglo-Saxon speaking countries.
In Italy it is roughly like this:
1. Up to the age of four, children go to nursery school (they play with them and do drawings or manual work).
2. Ages five to ten go to primary school (once called elementary school) for five years.
Until some time ago, ratings were given by adjectives such as: excellent, good, fair, sufficient, mediocre, ineffective. Today they use letters A to D (along with a verbatim report) but there are no more failures. In primary school today no one repeats the year. It is a compulsory school.
3. From eleven to thirteen years old they attend lower secondary schools (once simply called middle schools) for three years. The evaluation is made with numbers from 0 to 10, up to 5 you are insufficient, from 6 to 10 you are promoted. It is the same for everyone and is mandatory.
4. From fourteen to eighteen years old they go to a high school (now called secondary school) for five years. It's optional.
You can choose between various types which already lead towards subsequent humanistic, technical or scientific studies or towards a profession.
For example, there are: Scientific High Schools, Classical High Schools, Linguistic High Schools, Industrial Technical Institutes (electronic, electrical, mechanical, IT, telecommunications, nautical, etc. etc.), Technical Institutes for Surveyors, Commercial Technical Institutes, etc. etc.
The votes are numbers from 0 to 10 (accompanied by detailed textual explanations), up to 5 you are insufficient and from 6 to 10 you are promoted.
5. After graduating from high school you can enter university. The master's degree lasts a minimum of five years and you can become: doctor, engineer, architect, lawyer, physicist, mathematician, chemist, man of letters, psychologist, philosopher, etc. etc.

I would like to know how it works in your countries.

Actually, even better.
Consider only one engine.
Measure once with the reducer and once without.
What outcome do you foresee?

Sorry Roy, but to answer the question I asked you (two experiments hypothesis), consider two identical engines. Identical means equal rpm (before putting the gear reducer). Afterwards one will logically have fewer turns than the other. Same supply voltage (which we will not vary) and same power as the motor's data on the plate. In these conditions, in your opinion, which of the two motors will absorb more current, the one with the gear reducer or the one without?

Hi Alessandro. Agree with all your logic.
To get to the end first, that is a motor with and a motor without a gearbox.
The propeller in the water is the load. When driven direct you also need to know the revs per minute (rpm) of the prop.

With a gearbox in place same voltage etc, the rpm of the motor will say how much work is being done. The motor will be turning faster and if it now is able to get into its most efficient working area it will use less current. So compare the two propeller speeds.

Most motors will work with 30 % higher voltage as long as it reving highly. So if in the above case the gearbox has reuced the speed of the propeller then a small increase in battery voltage may get the speed back up and also be running more efficientl.

It is a case of monitoring the W = I x V.

These days I can make a good guess as to how much power is required looking at hull length weight and type of vessel. So for a cargo boat 1 metre long weight 6 or 7 Kgrms then 20 watts. It will run nicely on 10 - 12 watts but the 20 watts is for stopping quickly when there are other model boaters near.

I firmly believe that most boaters know where their models are, but not necessarily where they are going!
regards
Roy

Hi Roy.
First of all thanks for your answer.

I tried to reread your message several times but I'm not sure I understood everything. In fact, unfortunately, through my fault, I understood very little.
It often happens to me that, if I have to write from Italian to English, something more or less comprehensible may come out but incomprehensible translations come out from English to Italian.
Some key terms are not only mistranslated but completely misrepresented, making it impossible to translate the sentence correctly.


This time I was unable to understand the connection between some parts of the speech and the connection with the main question I asked you.

I understand that you helped your son with a physics test and that your son got an A+.
I guess A+ is the highest grade. Our system is different, it is not based on letters but on numbers or adjectives.
I won't write here how it works so as not to burden the discussion, perhaps I will do it in another message.
I can tell you, however, that I had no doubt that your son, thanks to you, would get a good grade.

Returning to my question, I wanted to submit this reasoning to you.
Let's see what you think.

It is true that the reducer is a gear system that has its own efficiency. Let's say 95% for example because with friction some of the energy will be lost.
However, by decreasing the rotation speed, the reducer will increase the torque. For example, if you halve the speed you double the torque.
Tell me if what I have written so far is correct.

Saying that the torque increases means that the engine requires less force to do the same work.
Is this correct?

Now I don't want to deal with the complex discussion of gears but I would like to draw a parallel (if it is possible) with pulleys.
Those who build the masts and running rigging, like us, are certainly familiar with these concepts and these examples.
That's why I use them.

1. A single fixed pulley cannot be considered a true lever. In fact, if I have to lift a 100 N weight, with a fixed pulley I will have to apply a force of 100 N (first image attached).

2. A single mobile pulley is a second class lever (I remember even if everyone knows that the second class lever is always advantageous). In fact, if I have to lift a 100 N weight, with a fixed pulley I will have to apply a force of 50 N (second image attached).

3. With two pulleys (compound pulley), one fixed and one mobile, you obtain the same result as in point number two (third image attached).

4. With two mobile pulleys, if I have to lift a 100 N weight, I will have to apply a force of 25 N (fourth image attached).


In reality we know very well that we will never be able to halve or divide the effort into four because there is friction.
The more postponements you make, the more friction dominates, so much so that it is no longer convenient beyond a certain number of postponements.

However, it is clear that a person using a hoist makes less effort than a person pulling the weight directly.
Here, I would like to compare the person to an electric motor and the gear system to the hoist.


So the question could be reduced to this experiment.
I take two identical electric brushed motors, to one I directly connect the shaft for the propeller and the propeller.
To another I first connect the gearbox and then the shaft for the propeller and the propeller.
I put an ammeter in series (paying attention to the motor plate data so as not to exceed 10 amps) and measure the consumption before one and before the other while the propeller turns in the water.
In your opinion Roy, but also in your opinion, which of the two will be the motor that consumes more current?
In my opinion (but I could be wrong) the one without a gearbox consumes more, because the effort will be supported entirely by the electric motor and will not be reduced by the gearbox.
But I'm curious to know your opinions on the matter and to see the result of the experiment.
If the amperage exceeds 10 amps (typical of the second fuse of normal testers) it is better to use a current clamp which however is much more expensive.

Alternatively, if the difference in amperes is too small to be able to establish who consumes more, we can do this:
I take two identical electric brush motors, to one I connect a spool with a wire and to another I first put a reducer and then the spool with the wire.
I have a weight lifted, attached to the wire.
The one without the reducer will be tested first and the current absorption will be measured.
Then the one with the reducer will carry out the test and the current absorption will be measured.
Same question.

I plan to do it as soon as I buy a gear reducer, two identical propeller shafts, two identical couplings and two identical propellers, I already have the identical motors.
Anyone who already has all the material can do this experiment and report the results to us, perhaps documenting them with a video.

Hi Alessandro re post 6.
A motor driving a gearbox is spinning faster so more friction on the brushes but less current but more cooling from the faster turning armature. It is more likely to be running in its most efficient set up.

The prop loading on direct drive should only be at an efficient voltage/load set up.
The technical comparison is difficult to do.

I helped my son do an A-level physics project many years ago which was to see the difference water pressure had on a spinning prop.

I designed an electronic rev counter and sensor which could be used under water. The prop shaft coupling had white paint on it and I stuck a piece of black tape that covered 180 degrees of the white surface. The sensor was an optical transistor or I may have rubbed the paint of an OC170 transistor. It also had a small torch light to get a good mark space waveform from it. It was 40 years ago now.

I had a direct read out on a meter, I think it followed that the deeper the water the slower the prop for a fixed power i/p. There was no cavitation but at the surface it was interesting and I wish at that time we had a surface piercing prop. But we also got to check on performance of the motor.

For this I made a Meccano frame and used a machined 1 inch diameter drum that locked on to a driven shaft. There was a belt over the drum that had a precision arm check instrument at the end of the belt. The instrument used to check relay contact pressures. I could wind this up with a screw to get a very precise reading.
We took measurements and also worked out effective brake horse power.
Ron Warring back in the 1950s in Model Maker magazine checked electric motors in this way and my figures were close to his.

There was a lot of calculation (no calculator then) and a graph was drawn with rpm and power i/p which told us the efficiency at different loads.

It was done just for interest and my 20% of stall current was good enough to find the best rpm.
But observation is easier and if the motor gets hot it is straining too much and especially so if the battery gets hot as well.

I did a restoration commission ex i/c drive and a Straight Runner, a few years ago and used an old Hectoperm motor (red and black) the fixed part was a 6 cell battery which also determined the esc. I got the 42 inch boat to plane and then cut back on prop diameters and pitch ( I have a lot) until it just backed off and then went back to the previous prop and it worked well on a bit less than 25 Watts (3 1/2 amps) and the battery lasted over 40 minutes.

She is Dolphin in my harbour, the new owners were very happy.

It is always worth doing a bit of maintenance as well. A slotted screw driver to get between the commutator segments and clean out deposited carbon and oiling the bearings. If you dissemble a motor removing the armature wil slightly degrade the permanent magnet. Taycol field wound magnets like oil on the brushes, but not much.
Not sure I can answer your question but thought I would let you know I do play with them quite a lot. It is the electric oily smell that I like.
By the way my son got an A+ for the work.
Roy

Hi Ronald, thanks for sharing the pdf, there are many useful and interesting indications.

I really liked this method, I find it simple and brilliant:

"Chuck the motor shaft in a drill press that runs at a known speed (e.g., 3600 rpm);
While running your motor as a generator, measure the voltage across the leads (maybe you measure 1.2 v);
Divide rpm by volts- in this example, 3600 / 1.2 = 3000. This is the approximate motor constant K, = RPM/V. Multiply K by desired running voltage to get free running speed. Example, at 7.2v, free running speed = K x V = 3000 x 7.2 = 21,600 rpm."

Hi Roy.

I agree with everything you wrote.
The stall current rule of thumb that you laid out is very interesting.
Maybe I didn't understand something because of the translation so I'll ask you directly:
In your opinion, all other things being equal, is a geared motor subject to less strain than a direct drive motor?
In my opinion certainly yes, but I want to listen to your opinion and your explanations which matter a lot to me.

Hi, not sure if this helps but here goes.
An unknown brushed electric motor, but one you might use for models, and assuming you have a multimeter that will read up to 20 amps.

There is a rule of thumb that if you apply the voltage you wish to use and you then stall the motor for just long enough to read the stall current. Then take 1/5th. or 20 % of this current reading and then when running normally, load the motor with a prop in the water at this current, this will be the most EFFICIENT speed.
Not the most power but where o/p and power consumption give the best option.

i.e. if stall current is 10 amps then run the motor at 2 amps. You can increase the load by changing the prop, either in diameter or pitch or both.

Frequently a gearbox which is free running and with a reduction of 1 : 2.5 is a good choice. If either the motor or the battery gets hot then the heat is coming from the battery supply and not doing anyone any good.

But do remember that the gearbox also places a load on the motor. So consideration should be given to this.

Generally a larger slower moving propeller where appropriate is more efficient. If it is speed you are after, you are trading off duration, make your choice and if the battery or motor get hot then have a re-think.

Cooling the motor is not a solution to the efficient use of the battery.

The heat from a brushed motor is centred around the brush gear where there is friction.
Given a choice of 2 motors one, 6 volt at 4 amps an a similar motor at 12 volts at 2 amps load then the 12 volt one is the most efficient. The Marx-luder motors often come in 2 versions. If you are not sure why then use Ohms law ascribing the same resistance values to the brushes.

regards
roy

More interesting items from the list.

https://matthewsmodelmarine.wordpress.com/2022/09/20/my-free-models-on-grabcad/

Chris I believe that by slowing the prop down by adding a gearbox you are actually running the motor at a higher RPM to achieve the same prop RPM.
Len

Very interesting and useful resource.

I'd read that running motors at slower speeds drew more amps and possibly caused motors etc. to run hot.

But in my Fairey River Cruiser I decided to do just that. Though I have used an oversized brushless motor, 35mm dia. in a 23" boat and tootling around everything stays cool. That motor is powerful and only driving a small 25mm prop so not working very hard at all.

The model is never going to get that much use either, so not worried about the longevity of the motor which I'm sure will be fine anyway, as it was more about building something unique and my faster Faireys will be used more.

Chris