ok I calculated the height/depth for a shaft housing a glass weight half the size (which is the optimal height for the weight in any shaft of fixed height, I as

ok I calculated the height/depth for a shaft housing a glass weight half the size (which is the optimal height for the weight in any shaft of fixed height, I assume you understand this is optimum, if you want a derivation I can provide it upon request) For clear skies (typical of deserts), we can use the daily insolation at sea level: 21,6 MJ/m^2 (I will ignore inefficiency of solar panel, the less efficient, the less deep or tall we need to store the energy generated, and one time construction of the storage is preferable over reconstructing the weights and shafts once more efficient panels become available) So if we want to store it as potential energy of a glass weight under the solar panel that generated the energy we have for each square meter of panel: a glass weight with cross section one scquare meter, and height l = L/2 where L is the height of the shaft. glass weights 2500 kg/m^3, so we have mass of the weight: m = l * 1 m^2 * 2500 kg/m^3 = l * 2500 kg/m = L/2 * 2500 kg/m We want the weight to be at all times contained in the shaft so it can be closed and sand can not blow under the weight or into the mechanisms. the height difference for the glass weigh at the top -but still completely inside the shaft- to the bottom -but also still completely inside the shaft- is h = L - l/2 - l/2 = L -l = L - L/2 = L/2 The potential energy difference for a weight of mass m and vertical travel distance h is E = m * g * h Plugging everything together: E = ( L/2 * 2500 kg/m * 9.81 N/kg * L/2 ) Solving for L = 2 * sqrt ( E / (2500 * 9,81 N/m)) Since the assumed insolation is 21.6 MJ/m^2 then one square meter of (ideal futuristic) solar panel will simply generate 21.6 MJ = 21,6 MNm Plugging into the equation for L we get L = 59,35 m I seeriously underestimated potential energy I guess, I thought it would have been larger. If your panels are less efficient (say 10%) or you have a large unused area you can of course make the structure less tall/deep.