GLYNN GUEST discusses a more analytical approach to your next model
In the course of my working life I’ve met some very intelligent people, often with a string of letters after their names to prove it. One thing that could occasionally amaze and to be honest also amuse me, was that some of these people would literally run away from any problems that required a calculation to produce a quantitative answer. There were also a few who could not cope with the idea of 'estimating' to obtain an approximate answer, which is handy to gauge the 'size' of something, or even determine if an idea is possible or not.
Looking at the questions that often get asked on Internet forums, sometimes over and over again, it struck me that many modellers seem unsure of how to calculate their way out of a problem. So what follows now, assumes that our struggling modellers know the basic mechanics of arithmetic, but may well have probably never learnt how to apply it to real world situations. Most of the time you just need to use addition, subtraction, multiplication and division in order to obtain a result. Well, squares, cubes and their roots will also come into it, but as these now only require you to press the right button on a calculator, quite a change from when I first learnt about them so they ought not to be a challenge.
The aim here is to use simple and straightforward sums so that you can get your next modelling project in, to use the phraseology of our American cousins, 'the right ballpark'. This could avoid disasters such as a model becoming too heavy to safely launch, or the opposite with a model too small to float when loaded with all its internal equipment. It is surely worth the effort of a few sums to avoid starting out on a difficult if not impossible new project?
Working out the scale of a model or how to change a plan to suit the size of model you want to build is a common source of confusion. To those skilled in such matters, it can seem puzzling that a simple Fraction that gives the Ratio that will produce the desired Percentage change is so difficult to understand, but therein may lie the nub of the problem as we have used three different terms for what is basically the same process.
It may cause no problems for the mathematically fluent to use such terms in this interchangeable fashion but it does, I know for a fact, cause some people a lot of confusion. Figure 1 is used to show that fractions, decimals and percentages can all mean the same thing. Taking '0' to mean nothing and '1' to mean everything on this scale it ought to be easy to see that the mid-point can be represented by the fraction '1/2' or if you prefer 'half'. Further subdivisions’ into it can be made such as the common fractions of a quarter (1/4) and three quarters (3/4).
Looking at these fractions and it is clear that the number on the bottom is how many equal parts the whole thing has been divided into. The number on the top is the number of these parts we are concerned with. There is no need to remember the terms numerator and denominator that math's teachers bandy about!
So, if we encounter the fraction of 7/16, all it means is that we have divided something into 16 equal parts of which we are going to need 7 of them. Hopefully this shows how we are going to use it when we try to make a model that is scaled down in size from the original object.
I have to confess to a preference for using a fraction of 1/144 when building radio control model warships. It might seem to be an odd fraction in this decimal age, but is related to the British Imperial measurements and equates to one inch equals 12 feet.
So, if I’m planning a new model then the original vessels dimensions would have to be reduced to 1/144 of their value to make the model. This is easily done with the aid of a calculator by simply dividing by 144.
Sometimes the fraction is not so simple and your model might need to be built to a scale of say, 14/78 to make a practical size. This would entail dividing all the original sizes by 78 and then multiplying them by 14. This is two operations and being lazy, I prefer to do just one. This can be achieved by dividing 14 by 78 to produce 0.179 (yes, I have rounded it off, figuring that a few ten thousandth's can be missed!) then placing it the calculator's memory to allow it to be used over and over again. Now all you have to do is multiply the original’s dimensions by this number to get any size you need for a model at this scale.
This leads neatly onto the idea that any fraction can be represented by a decimal figure. Thus referring back to Figure 1 again, we can add a decimal scale so that the fraction 1/2 is equivalent to the decimal 0.5; a 1/4 becomes 0.25; and so on.
Percentages are often quoted when trying to scale things up or down. Now the scale reads from 0%, which equates to our 'nothing' and 100 % which is 'everything'. Half on the percentage scale become 50% and three quarters (3/4) becomes 75%, and so on, from which it should be clear that the percentage value is just the decimal value of any fraction multiplied by 100.
Now whenever someone talks about terms such as 3/8 or 0.375 or 37.5 %, they all mean the same thing. You can use whichever term you find most acceptable. I like decimals as they require one multiplication when using a calculator, but for mental arithmetic, fractions always seem easier to work with.
Quite often a modeller sees a plan or drawing of a vessel that they would dearly love to build, but it’s the wrong size. A common example might be a plan in a magazine or book and unless you like massive page sizes, it is going to be way too small for a practical working model.
The simplest case is where you just decide how big you want the model to be, perhaps 750mm long, and then measure the length on the plan, so let’s say 90mm in this example. So now we know 90mm on the plan equals 750mm on the model. Thus, dividing 750 by 90 gives us the conversion ratio for scaling-up any dimension on the plan to match a model of this proposed size. Just put the ratio of 8.333 in this example, into a calculator’s memory and off you go, as each 1mm on the plan equals 8.333mm on the model.
It is a little harder if you want to change a plan or drawing made to a certain scale into a model built at a different scale. Here is where it can become confusing with fractions, ratios and percentages all getting entangled, but the actual process is straightforward. You could print the original scale drawings, enlarge them to full-size, then reduce them to the desired scale size! Luckily we do not have to actually do this, as imagine trying to get a piece of paper to match the dimensions of the full-size vessel!
By using a little mathematics, this process can be simplified to the following equation:
Model Dimension = Plan Dimension x desired Model Scale/original Plan Scale
The last two terms, Model Scale and Plan Scale, are going to be constant for all the dimensions we are converting, and so might as well be turned into the decimal ratio and put it into a calculator's memory for easy repeated usage. Plan and model scales are usually quoted as fractions, so an example could be to convert a 1/50 plan into a 1/32 scale model:
1/50 = 0.0200 and 1/32 = 0.0313
So the desired conversion ration is 0.0313 divided by 0.0200 = 1.5625
Yes, there are other ways to do this calculation, but until you are fluent in the process and have a feel for what the numbers ought to be, this is probably the safest way.
Altering the linear dimensions of a model, such as the length and beam etc. can have more far-reaching consequences than some modellers might expect. This is often shown by the person who reduces the size of a plan to make a more convenient model and finds that it is now unable to carry the weight of batteries and radio gear. The converse can also occur in that a modest increase in size of a plan may produce a monster of a model that is a pain to handle.
This is the infamous 'Square-Cube' relationship that can perhaps be best understood by considering what happens when you double the size of a simple cube as in Figure 2. Starting with the original cube, let it have edges of one unit length (this unit can be a centimetre, inch or whatever takes your fancy). The volume of such a shape is found by length x breadth x depth. So, our unit cube has a volume of one cubic unit of whatever you used for the length of the sides.
The cube has six sides each made of a square shape. The area of each side is the product of the length and breadth of the sides. Thus, the unit cube has a side area of one square unit giving it a total area of six square units.
Let’s now see what happens when we double the size of our cube so that it has an edges two units long. The area of each side is now four square units and with six sides it gives a total of 24 square units. This is a fourfold increase compared with the original cube size. If you thought that was a big jump, then consider the double size cube’s volume, as it has actually increased by eight times!
By playing with other increases in cube size from the original unit cube, it is not hard to see that areas change by the square of the conversion ratio you use and the volume by the cube of this ratio. In the previous example the square of 2 (i.e. 2 x 2) is 4 and the cube of 2 (i.e. 2 x 2 x 2) is 8.
The same effect is found if instead of enlarging the cube, we reduce its size. Using Figure 2 again, lets start with the big cube and halve its size. The surface area is reduced from 24 to 6 square units, a quarter of the original size and the volume shows an even larger reduction back to one eighth of the double size cube.
The same rules apply as before, but because the conversion ratio is less than one (it's 0.5, or if you prefer 1/2 in this example) then things get smaller. Square this ratio (1/2 x 1/2) and you get the 1/4 of the original area; cube it (1/2 x 1/2 x 1/2) and you have one eighth of the volume.
You might think that all of this does not really matter if you are changing the size of a model plan by a mere 25%, but scaling-up will more than double the volume and hence its weight. Scaling down will perhaps be more of a problem as the model can only support a shade under half its original weight. Table 1 lists the effect of different conversion ratios on Volumes (i.e. weight) and Areas.
You might be wondering why 'area' is so important in our models? So consider a yacht being scaled down where the sideways force on the sails is being countered by suitable ballast weights? The force on the sails depends upon the sail area and if the new model is to be half the size of the original, we can expect the force to be a quarter of the original. However, the model's displacement will now only be an eighth of the original and thus greatly increasing the chances that it will get blown onto its beam ends when sailing. Yes, this an example full of simplifications, but it shows that reducing a model design in size is not as straightforward as at first you might think. Even scale models can suffer from the effect of side winds!
It is not a bad idea to work out, if only approximately, the likely weight of your next project. As shown previously, it is a straightforward process to calculate the effects of changing size of your model if you are starting from a design of known weight, but what if it is an original design you plan to build? It may be possible to use other model's designs for guidance, but this requires them to have a similar hull shape and size and this is not going to happen for every new design you plan to build.
If it is to be an exact scale model, then you could use the displacement of the full-size vessel to calculate the model's weight using our knowledge of the effects of scaling model plans. In this case you take the full-size weight and multiply it by the CUBE of the scale you plan to use for your model.
For example: If the full-size vessel displaces (weighs) 4000 tons (or tonnes as there is not much difference) and you plan to build a model to 1/100 scale then:
The cube of 1/100 is 1/100 x 1/100 x 1/100 = 1/1000000 (i.e. one millionth!)
Multiply 4000 by one millionth (or divide by one million for the same result), this gives you 0.004 tons, which I agree is not a very convenient figure with which to build models. However, since one ton contains 2240 pounds, just multiply this value to your model weight in tons to get 8.96 pounds, or for those of you with a leaning towards the metric, 1 ton(ne) contains 1000kg so our model will be approx. 4 kilograms in weight.
One thing to watch out for when using the weight of a full-size vessel is that the appropriate value is used. A variety of tonnages can be quoted, but what you need is the one that matches the waterline at which you would like to float your model.
All of the preceding is of little value if you are building a model that is semi-scale or freelance, or in other words, you have no real full-size vessel that matches your proposed model. True, you could try to find something that closely resembles your idea, but this will not work if you have simplified the hull shape for easier building, and perhaps increased the draught and beam of the model, perhaps for better sailing qualities.
Luckily, we can still make a good estimate of a model's weight by using the fact that they float as soon as the hull has displaced a volume of water with the same weight as the model (Archimedes' famous principle). This of course simply requires you to measure the underwater volume of a model's hull and then multiply it by the density of the water. This would be easy if we made our model hulls with simple shapes, such as rectangular and then its volume would be length x breadth x depth! Alas, our hulls have to pay some attention to streamlining, or at least they ought to, which results in shapes that are anything but simple. There are ways to calculate the volumes of such a hull, by slicing it into numerous transverse cross-sections and working out the area of each section, multiplying it by the thickness of each section and finally adding them all together. This is quite accurate although tedious, but thankfully we have an easier way that will produce an approximate value that is usually good enough.
We can use the idea that a hull, full-size as well as model, only occupies a fraction of the block created by multiplying the length (strictly the waterline length) by its maximum beam and draught, Figure 3. Within reason, the shape of a vessel's hull is determined by its function; for example warships tend to have slim hulls whilst others such as tugs can be stouter in shape. Thus, many warships will tend to occupy a similar fraction of this block whilst the fraction for tugs will cluster around a different value.
This fraction is usually termed the Block Coefficient of the hull. I have seen values quoted for full-size vessels, so please see Table 2, and you could use these as follows:
Model Length x Beam x Draught x Block Coefficient x Density of Water = Weight
For example, consider a hull having a Block Coefficient of 0.5, a model length of 75cm, beam of 10cm and draught of 5cm, the density of water being a convenient 1g/cc (one gram per cubic centimetre)
75 x 10 x 5 x 0.5 x 1 = 1875g (or 1.875kg)
Or, for the diehard Imperialists amongst you, for the same Block Coefficient the length is 30 inches, beam 4 inches and draught 2 inches, the density of water now being a less convenient 0·58oz per cu. inch.
30 x 4 x 2 x 0.5 x 0.58 = 69.6 ounces or 4.35 pounds
My usual method of working out the likely weight of a new scale model is based on experience gained with previous models. Working backwards from the known operating weights of these models, it was found that if this weight (in ounces) was divided by the overall length x beam x draught of the model (in inches) the result came out around the fraction of 3/8 (or 0.375). So I tend to use the following:
Overall Length (inches) x Beam (inches) x Draught (inches) x 3/8 = Weight (ounces)
When I discovered this handy method, most of my models had been based on warships which had similar slim hull shapes. So, using it for more portly hulls does tend to underestimate the model's final weight, but this is probably safer than the opposite, which can lead you into trim and stability problems.
You might like to try this method, in which case you need to determine your own fraction. It will depend on the type of models you build and the units you use. It will never give you a precise answer, but at least it will get you in 'the right ballpark'.
Power and more power?
When building a scale model it is possible to take the power of the full-size vessel and attempt to scale it down to suit the model. This method can be full of pitfalls, and such results will always be suspect until tested. For this reason, most modellers use past experience and then err on the generous side. Their very sensible reasoning being that with a little too much power you can always 'throttle down' and save the excess for emergencies.
The actual relationship between a model's speed and the power available is not the simple 'double the power for twice the speed' that some people might imagine. By the way, that which follows, refers to displacement models and not the planing types which achieve their high speeds by using hydrodynamic lift rather than just pushing the water out of the way.
It is a sad fact that if you want to double a model's speed then it will need approximately eight times the original power. This is our old friend, the Cube Law that we encountered when discussing the effects of scale, i.e. doubling the speed needs 2 x 2 x 2 = 8 times the power, which is shown in Figure 4.
In terms of a formula, the relationship can be written as:
P = k V cubed
Or in words; Power (P) = Constant (k) x speed (V) x speed(V) x speed(V)
This constant depends upon the shape and size of the hull. Slim low resistance hulls will have a small constant, whereas bluff angular hulls result in larger constants. Likewise, smaller hulls have smaller constants.
From experience with many models I have found that destroyer type hulls built in 1/144 scale with a length of around 30 inches (75cm), a good value for k is 0.10. Thus, if the model is required to sail with a top speed of four feet per second, then the power required is:
P = k V cubed, which equals 0.10 x 4 x 4 x 4 = 6.4 Watts
At this scale, larger warship models such as cruisers will have k values of 0.15 to 0.25 and battleships and aircraft carriers have k values of 0.35 and larger. It is worth pointing out that the value of k is affected by how well the components in the drive-line are set up and matched. Poor alignment, high friction and the wrong motor/propeller combination will all serve to increase the power needed at any model speed. Also, the numerical value of k depends upon the units of power and speed used. I started many years ago with Watts and feet per second, and do not feel like changing!
It is also interesting to see that increasing a model's speed will result in a disproportionate increase in the power demanded, but reducing the speed greatly lowers the power demand. For example, consider a modest speed change of 20%:
A 20% increase requires the power increasing by 1.2 cubed =1.73 or approaching double.
20% decrease requires the power decreasing by 0·8 cubed = 0·51 or almost half.
So, easing-off on the throttle stick can significantly increase the running time of a displacement model, which is something that might seem obvious, but running at 85% of full-power speed can markedly increase sailing times.
Exactly how a new model will respond to the transmitter controls is always something of an unknown! Luckily I have kept records of my models and after sorting though them it was clear that the power to weight ratio was the best indication of a model's sailing character. The results are shown in Table 3 for these displacement models.
The STEADY models were exactly that, relaxing to sail and would never get into dangerous situations. The HANDY category had a little more sailing character, being more nimble, but never stressful. Those classed as LIVELY had a good combination of sailing qualities, being responsive but requiring more care for precise manoeuvres. Full attention was needed when sailing the EXCITING models, as coarse operation of the transmitter controls could get me into trouble!
For the few fast scale craft I’ve built, torpedo boats and the such like, a figure of around 20 Watts/pound (40 Watts/kg) would give the right sort of performance at 1/32 scale, but like most things, you need to establish your own values for comfortable power to weight ratios. What works for me with my models on the waters that I sail on and in the style that suits me, might be quite different for you.
Hopefully, I have shown that performing a few calculations before embarking on your next project can place it very much in the 'doable' region of our hobby. That is to say, the size of the model will neither be too large to make launching and recovery an injury risking process, nor will it be too small to float when outfitted with your intended equipment. Likewise, the performance of the model will be about right. A little tweaking might be needed to get it perfect, but you ought to avoid having to replace the entire driveline to get what you want.
This approach is admittedly full of approximations and assumptions rather than an exact science. The alternative is to go down the route of a frustrated naval architect and try to create a perfect design on paper or a computer screen, but I suspect that few such perfect designs have ever survived unchanged after contact with the harsh real world!
Rather than spend a lot of time, effort and money (all of which are better employed in building models) whilst aiming for perfection, I’m happy to check that the new model is in the doable zone. This approach has not failed me yet, although I will confess that a few projects have caused some head-scratching when they have come a shade too close to the boundary between doable and impossible, but then life would be awfully boring without a few such challenges!
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