# How do I calculate new minimal speed for a glider when I put water into the tank?

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?

• If the water tank(s) are original equipment on the glider, you will always find the answer to this in the operating handbook. – Juan Jimenez Aug 16 '18 at 19:10
• That's clear, but doesn't answer my question... – michael Aug 19 '18 at 11:37
• 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 ?? – Juan Jimenez Aug 21 '18 at 11:42
• Yes, that's wrong. Your calculation is correct. Where did you find the equation you refer to as being "in the theory"? – Finbar Sheehy Sep 13 '18 at 15:06
• 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... – xxavier Sep 13 '18 at 15:37

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