§ 17
How much solder should I use for 20 pipe connections?
Answer: In hard soldering, the solder is drawn into the gap between the two parts (capillary action). The amount is therefore the gap's volume: length × overlap × gap width, plus a little for the small fillet that should stand at the edge. The calculator gives grammes per joint, in total and what length of wire it corresponds to, so you can cut the pieces off beforehand and order correctly. The gap should be where the solder's data sheet says, typically 0.05–0.2 mm.
Spaltens rumfang = længde × overlap × spaltebredde (mm³)
Lod (g) = rumfang × massefylde ÷ 1000 × (1 + tillæg) (1 g/cm³ = 0,001 g/mm³)
Trådlængde = rumfang × (1 + tillæg) ÷ trådens tværsnit π × d²/4
The joint should lie within the plumb's direction, typically 0.05–0.2 mm. Too large joint does not fill, too small joint presses too hard on the edge plate or creates bonding force that holds the web in place during forming.
| Symbol | Meaning | Unit |
|---|---|---|
| laengde | Joint length (around a pipe: π × diameter) | mm |
| overlap | Overlap (how far the parts go into each other) | mm |
| spalte | Column width | mm |
| antal | Number of assemblies | stk |
| rho | Solder density (datasheet; silver solder typically 8.9-9.6) | g/cm³ |
| tillaeg | Allowance for hollow-core and spillage | % |
| traad | Plumb line diameter | mm |
| = | Lod i alt | g |
| = | Solder per connection | g |
| = | Wire length per joint | mm |
| = | Total wire length | m |
Does not apply when
The calculator's own default values inserted. The figures are calculated with exactly the formula the tool uses.
Substituted
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Når du skal hårdlodde en serie samlinger, kan du regne loddet ud og klippe stykkerne af på forhånd. Det giver ens samlinger og mindre spild.
Du skal lodde 20 samlinger på Ø15 kobberrør med 11 mm indstik og 0,1 mm spalte.
It means: Du skal bruge 11.4 g lod i alt, 0.57 g pr. samling. Klip stykker på 35 mm af tråden.
Seks lampefødder skal have bunden loddet i, med 314 mm omkreds og 4 mm overlap.
It means: Det kræver 8.3 g lod, svarende til 0.51 m tråd.
En samling på Ø15 rør tager ca. et halvt gram sølvlod.
Rumfanget i mm³ × massefylde ÷ 1000 skal give grammene.