## Can you check my sourdough percentage math?

I'm not new to percentages, but I'm new to sourdoughs. I don't know why my brain won't wrap around this so if you guys can double check my math I would appreciate it.

I want to make a sourdough at 70 percent hydration with 20 percent of the total dough coming from the starter. The starter is 100 percent hydration . 2% salt....

I have the following percentages...

Flour 551

Water 385

Starter 240

Salt 24

The final dough should be around 1200 grams for two 600g boules. I got these numbers by setting the sourdough to 20 percent of the final dough weight. Salt came from the same. I took the remainder of the dough weight, i think was 936 or something, and set my flour and water to that keeping in mind the starter was 100 percent hydration.

Am I way off?

If your starter is 100% hydration, then it's composed of 120g flour and 120 water. If so I calculate your overall water as 120+385 = 505g, and your total flour weight is 120+551 = 671g. That yields an overall hydration of 75%. Your salt, however, at 24g is nearly 3.58% which is high. Salt is typically around 2% of total flour weight.

So you're close, but need to tweak hydration down and salt down.

Hope that helps.

Larry

Hey thanks larry. I'm on the right track. I wrote the 24g wrong. I used .02 x 1200 twd instead of 2 percent of the flour weight. Good looks.

Final dough weight = flour + starter + water + salt

Let the amount of pure flour you add be F(i) (Flour-initial)

Let the amount of pure water you add be W(i) (Water-initial)

1200g = F(i) + Starter + W(i) + Salt

You want your starter to be 20% of the final dough, which is 240g. (1200 x 0.2 = 240g)

Of this you have 120g flour and 120g water.

Let the total amount of flour in the recipe be F(t), which will be F(i) + 120g

Let the total amount of water in the recipe be W(t), whihc will be W(i) + 120g

So now we have ...

1200g = F(t) + W(t) + Salt

Salt is 2% of the weight of the flour, or 0.02F(t)

Total water (W(t)) in the recipe is 70% of the weight of the flour, or 0.7F(t)

So now we have ...

1200g = F(t) + 0.7F(t) + 0.02F(t)

1200g = 1.72F(t)

F(t)=698g

So the amount of pure flour to add (F(i)) is 578g (698-120=578)

The total amount of water in the reicpe is 0.7F(t), or 489g. 120g of this comes from the starter, so the amount of pure water to add (W(i)) is 369g (489-120=369)

The total amount of salt to add is 0.02F(t), or 14g (0.02 x 698 = 14)

Thus, the final formula for your bread would be ...

1200g dough = 578g flour + 240g starter + 369g water + 14g salt

(it actually adds up to 1201g, due to decimal rounding errors)

GAPOMA, great response....thanks for taking your time out.

I've written it all into an excel spreadsheet. If you guys don't make taking a look and telling me if it makes sense to you I would appreciate it.

https://docs.google.com/spreadsheet/ccc?key=0AhFfnVHZTn2pdEptcl9WajNxb1IzS2lteVFmcF95Vmc