So I wanted to print a 3D parabola, with a radius of 15 cm and a height of 4 cm. I've told this company, they told me that they need 3.5 days and its price will be 147 USA dollars and 50 cents.

I just want to know what is the average price for this?

  • $\begingroup$ What material is this to be printed in? Are you supplying the model or is the company creating the model? There are other variables to consider to be able to supply a sufficient answer. $\endgroup$ – agarza May 18 at 20:56
  • $\begingroup$ company created model itself $\endgroup$ – math boy May 18 at 21:18
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    $\begingroup$ As a "Math boy" you would expect you do the math 😉. If it is a plastic model, slice the model, note the filament length needed and deduce how many is on a spool at what price and add some extra for your time, energy and depreciation charge of the printer. I am guessing the object to be printed is in metal considering the price. This question is unanswerable without more details, please supply details by edit. Thanks. $\endgroup$ – 0scar May 18 at 22:17
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    $\begingroup$ 0scar, you have listed T&M costs only. Of course the designer's work adds to this. And note company overheads (like premises, management, communications, ...), and packaging, shipping, payment handling, and the taxes. Thus it would be cheaper when ordered on bulk ;) $\endgroup$ – octopus8 May 19 at 4:03
  • $\begingroup$ Normally one thinks of a parabola as 2D. Are you thinking of the shape of a parabolic mirror on a reflecting telescope? $\endgroup$ – Perry Webb May 19 at 13:30

The price dependss roughly on material, machine hours, operator labour, profit and administrative overhead. Some companies deduce the operator labour, machine hours and overhead to roughly 10 times the material cost. I think that is kind of fair. In your case I assume that you use PLA, the perabola is hollow (just a flat surface) and the company needs to construct the 3D model the parabola to a specific tolerance. this model in PLA should not cost more than \$5 in filament, so about $ 50 should be an okay price for your parabola. The total price of \$147 leaves about \$97 for modeling the parabola. I think that's fair, given that a non-mathician has to find a way to construct a model and test the results. If, for example, you want a metal print and you provide the 3D model I find the price of \$147 to be very normal. These metal printers are very expensive, labour intensive and eat up energy. You can get instant quotes at the following sites: https://xometry.eu/, https://formalize-am.com/ All above is just my best guess, given the data you provided. Expect a better answer when you provide more data.

  • $\begingroup$ yeah company told me that they use PLA, is there any cheaper material? $\endgroup$ – math boy May 19 at 8:49
  • $\begingroup$ Not really. PLA is about the cheapest stuff you could get. But the material is not the most expensive. having someone do the labour is the expensive part. Still assuming you have to get the parabole modeled by someone, you could try to find an online mathematical model generator or download a parabola from Thingiverse. Some models are in openscad, where you can easily code a math shape. If you'd live in the EU I could help you out printing it. $\endgroup$ – Hacky May 19 at 8:58

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