Regards,

Peter Goedhart

Flits & Flash

Indeed my simplified formula is not meant to be accurate close the hyperfocal distance, it is meant for mental calculation, in order to quickly estimate the DoF while shooting a portrait, without to use an excel sheet.

When you rely on an excel sheet there is no need to simplify the formula in my opinion. ]]>

With 5000 in s and with :

Dn : =(C7*C2^2/(C2^2+C3*C4*(C7-C2)))/1000

Df : =(C7*C2^2/(C2^2-C3*C4*(C7-C2)))/1000

Result is :

Dn : 3.0241

Df : 14.4259

Good. Thanks for your help.

But, warning with the simplified formula. When s is very very near of hyperfocal, results are delirious !

Example :

f = 50 mm

A = 8

c = 0.03 mm

H = 10.42 m

If s = 10 m

Simplified : Dn = 250 m

Exact : Dn = 223.2 m

If s = 10.41 m (0.01 m add)

Simplified : Dn = 16265.62 m -> Delirious !

Exact : Dn = 1913.6 m

If s = 10.415 m (0.015 m add)

Simplified : Dn = 65093.75 m -> Very delirious !

Exact : Dn = 2099.79 m

For this reason, I use always a modified simplified formula for Df and Dn :

Df : (H*s)/(H+(s-f/1000))

Df : (H*s)/(H-(s-f/1000))

Now I know the use of exact formula, I understand wich the results are very similar at my modified simplified formula. My simplified formula are more modified with the case where d > H :

Df =IF((B5*B7)/(B5-(B7-B2/1000))<=0;"Infinite";(B5*B7)/(B5-(B7-B2/1000)))

Dof =IF(B8="Infinite";"Infinite";B8-B6)

Where :

B5 = H

B7 = s

B2 = f

B6 = Dn

B8 = Df

Oups for my english, I'm french user…

]]>C2 : f

C3 : A

C4 : c

C7 : s

Dn : =C7*C2^2/(C2^2-C3*C4*(C7-C2))

Df : =C7*C2^2/(C2^2+C3*C4*(C7-C2))

Thanks…

]]>Example for 50 mm Ã f/11, c = 0.03, s = 5 m :

With simplified :

Dn : 3.02 m

Df : 14.43 m

Dof : 11.4 m

With exact :

Dn : 4.97 m

Df : 5.03 m

Dof : 0.06 m

(make with OpenOffice Calc)

What’s the matter ?

Thanks…

]]>We checked and changed the numbers in the set.

Regards,

Peter Goedhart

Flits & Flash