Tech/maintenance Mathematics Question

[jbtsax]

I didn't know where to post this, so moderators move it to where it fits. If it doesn't fit anywhere I can ask the question as a personal message. This is primarily for the members who are mathematicians, but anyone is welcome to respond

Here's my challenge: I have been given the "harp" of a Fender Rhodes electric piano to gold plate the sound bars. To calculate the cost requires knowing the entire surface area to be plated. The bars are in 3 distinct sections, each of which has a slightly different shape.

The section with bars #54 to #73 (the smallest) has 20 bars that increase in length by 1.17 mm. All are 11 mm wide. The section with bars #26 to#53 has 28 bars that increase in length by 3.2 mm. These are 12.25 mm wide. The section with the largest bars #1 to #25 has 25 bars which increase in length by 4 mm. These are also 12.5 mm wide.

My question then is there a simpler way to calculate the surface area of each section of bars without doing them all individually and then adding to get the total? Is there some way to calculate averages than can work? Included in the total will be the area around the perimeter of the bars which run between 3.0 mm, and 3.25 mm thick. Any help will be appreciated.

 

[saxyjt]

You're talking about width of the bars, but what are the shapes? Round, square, anything else? That would help calculating the circumference that multiplied by the length of each bar will give us the area!

 

[nigeld]

I think the formula you need is this:

What this gives is the sum of k values, the first of which has value a, and where the size increases by b each time.

So, for example, suppose you need to add up 11 values, where the smallest is 2.1 and they increase by 1.3 each time, the total is

11x2.1 + 55x1.3

Edit: this is too complicated for what is needed - see @kevgermany post below.

 

[saxyjt]

I think the formula you need is this:

What this gives is the sum of k values, the first of which has value a, and where the size increases by b each time.

That's very clever and it looks great, but I can't make the result being an area. Is it?

Did you forget to multiply each occurrence with the perimeter of the related section?

 

[kevgermany]

All you need for this is to take the length of the middle bar in each section and calculate the area of that. Then multiply by the number of bars in the section.

To be safe, where you've got an even number of bars take the length of the bigger of the two. And use the biggest thickness throughout. Assuming the bars are rectangular cross section, you need to calculate the size of three sides and double the total.

If the bars are round the area of the curved surface is pi*diameter*length, the area of each end is pi * the square of the radius.

 

[nigeld]

The section with bars #54 to #73 (the smallest) has 20 bars that increase in length by 1.17 mm. All are 11 mm wide.

That's very clever and it looks great, but I can't make the result being an area. Is it?

Did you forget to multiply each occurrence with the perimeter of the related section?

My formula can be used to calculate the total length of a set of bars. This is then multiplied by the perimeter of the cross section to get the surface area.

So, to take the example from @jbtsax above.
There are 20 bars. All are 11.0 mm wide. The text does not give the length of the shortest one, so for the sake of the example, I will make up a number -13.7mm. Also for the purpose of the example I will assume that these bars are 3.2 mm thick.

Assuming that the cross section is rectangular, the perimeter of the cross section is top+bottom+two-sides = 11+11+3.2+3.2 = 28.4mm

Using the formula, we calculate the total length of all the bars L = (20x13.7 + 190x1.17)

If the ends of the bars are to be plated, then we must include them. There are 20 bars, each with 2 ends, each with area 11x3.2

So the total surface area of all the bars is the cross section perimeter times the total length, plus the surface area of the ends of the bars
= 28.4x(20x13.7 + 190x1.17) + 20x2x11x3.2

 

[nigeld]

@kevgermany 's solution is simpler than mine but gives the same result. I got carried away by the desire to remember the formula.

Another easy way to get the average length of a set of bars is to add up the lengths of the shortest bar and the longest one and divide by 2.

 

[nigeld]

Are these bars producing sound?
If so, the ratio of the lengths will presumably be logarithmic rather than linear.
(Much more exciting from a mathematical point of view!)

 

[kevgermany]

Are these bars producing sound?
If so, the ratio of the lengths will presumably be logarithmic rather than linear.
(Much more exciting from a mathematical point of view!)

They were measured. Linear increases in each set - see the first post.

 

[saxyjt]

I had no idea what these sound bars looked like, so I searched and found this:

Gives an idea what's at stake here! Quite a task. 🙂

 

[Alice]

I had no idea what these sound bars looked like, so I searched and found this:

Gives an idea what's at stake here! Quite a task. 🙂

I had an idea of what they were but that video has properly educated me! Wow!

 

[nigeld]

Judging by the video, the increases in length do not appear to be linear. If this is the case, then the ways to calculate the average length given by @kevgermany and me are not correct because the series is probably geometric (1 + a + a^2 + a^3 . . .) rather than arithmetic (a + 2a +3a + 4a ...).

@jbtsax - please could you let us know whether the lengths increase by a constant amount or whether the lengths increase according to the pitch. It is easy enough to calculate the average length in either case, but the result is different.

 

[jbtsax]

Judging by the video, the increases in length do not appear to be linear. If this is the case, then the ways to calculate the average length given by @kevgermany and me are not correct because the series is probably geometric (1 + a + a^2 + a^3 . . .) rather than arithmetic (a + 2a +3a + 4a ...).

@jbtsax - please could you let us know whether the lengths increase by a constant amount or whether the lengths increase according to the pitch. It is easy enough to calculate the average length in either case, but the result is different.

Thanks for your input everyone. @nigeld your formula is like Egyptian hieroglyphics to me without knowing what each of the letters represents. Can you fill that in for me?

I started by measuring the lengths of the smaller bars with calipers and quickly found out the increase in size varied by a small amount. Therefore I measured the shortest and longest in each run subtracted to get the difference and divided that by the number of bars in each run minus 1. That got me a good approximation of the average increase in length.

In a Rhodes it is the "tines" that are struck by the hammers and vibrate next to a pick-up that makes the sound. The attached "tone bars" simply add resonance through sympathetic vibration.

 

[nigeld]

Thanks for your input everyone. @nigeld your formula is like Egyptian hieroglyphics to me without knowing what each of the letters represents. Can you fill that in for me?

I started by measuring the lengths of the smaller bars with calipers and quickly found out the increase in size varied by a small amount. Therefore I measured the shortest and longest in each run subtracted to get the difference and divided that by the number of bars in each run minus 1. That got me a good approximation of the average increase in length.

In a Rhodes it is the "tines" that are struck by the hammers and vibrate next to a pick-up that makes the sound. The attached "tone bars" simply add resonance through sympathetic vibration.

OK. If we assume that each increase in length of the tone bars is more or less the same within a section/run, then the lengths of the bars form what is called an arithmetic series, which is what @kevgermany and I had initially assumed. A simple way to calculate the average bar length in the run is to add together the lengths of the shortest and the longest bars in the run and divide by 2.

That's all my formula says in a much more complicated way - I just got carried away with the sums. 😳

If the tone bar lengths had been tuned to pitches, then the way to calculate the average length would be different.

However, the safest way to calculate the average length of the bars in a run is to add up all the lengths of the bars in the run and then divide by the number of bars. It might be worth doing this for the smallest run to check whether it is more-or-less the same as shortest+longest/2. If it isn't, then I'm afraid you are going to have to measure all the bars.

If all the bars in a run have the same cross-section (i.e. same shape, width and height, but different lengths) then you can calculate the total surface area by calculating the surface area of the average-length bar and multiplying by the number of bars.

Edit: I have replaced "section" by "run", since it seems more obvious what it means.

 

[Jazzaferri]

As mathematical precision appears to not be required using a methodology that approximates the area of an average piece multiplied by the number of pieces will be close

 

[kevgermany]

We're being too clever. Lay the pieces from each section end to end, measure total length. Take biggest piece, measure circumference. Multiply. Store. Calculate area of one end in section using biggest piece. Multiply by number of pieces and again by 2. Add this to stored value to get plating area for first section. Repeat for remaining sections.

 

[Ivan]

The attached "tone bars" simply add resonance through sympathetic vibration

The additional weight of gold will detune the tone bars by a factor of what?

 

[kevgermany]

It's not just mass, but also the change in dimension.

 

[jonf]

Use Excel. Linear increases in each section. Calculate for the smallest and second smallest in each section. Excel will do the rest for you.

 

[jbtsax]

The project is finally completed. Due to the fact that I also gold plated the screws and washers that attach the bars I decided to estimate the amount of gold used by the rate of deposit in milligrams per amp minute. It came out to 2.5 grams the amount in 1 liter of gold plating solution that costs $325.00. I do have the benefit of the mathematics I learned from you folks to apply to future projects, so thank you again for your help.