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Lets say a glider weights 400kg and has a minimal speed of 70km/h.

When it has additionally 100l water, so in total 500kg, what is the new minimal speed?

I would argue, that in order to reach the same point in the polar diagram, we have to scale according to

$v_2 = v_1 \cdot \sqrt{m_2/m_1}$

because drag and lift forces scale quadratic with speed.

The result would be 78km/h.

However, in the theory we learn

$v_2 = v_1 \cdot m_2/m_1 = 87km/h$

Q: Isn't that wrong?

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    $\begingroup$ If the water tank(s) are original equipment on the glider, you will always find the answer to this in the operating handbook. $\endgroup$ – Juan Jimenez Aug 16 '18 at 19:10
  • $\begingroup$ That's clear, but doesn't answer my question... $\endgroup$ – michael Aug 19 '18 at 11:37
  • $\begingroup$ Fair enough. So let's go back to the beginning. What do you mean by "minimal speed"? I am a glider pilot and we don't use that term. Do you mean stall speed? Or ?? $\endgroup$ – Juan Jimenez Aug 21 '18 at 11:42
  • $\begingroup$ Yes, that's wrong. Your calculation is correct. Where did you find the equation you refer to as being "in the theory"? $\endgroup$ – Finbar Sheehy Sep 13 '18 at 15:06
  • $\begingroup$ Lift is related to airspeed, all other things equal, by v squared. Hence, your calculation is OK, and that 'in the theory' is clearly wrong... $\endgroup$ – xxavier Sep 13 '18 at 15:37
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$v_2 = v_1 \cdot \sqrt{m_2/m_1}$ is the correct formula.

$v_2 = v_1 \cdot m_2/m_1 = 87km/h$ is simply wrong, possible that it is only a typing mistake, but it is wrong just the same.

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