I can assure you that New Zealand is renowned for its lightweight pie crust, and regardless of the use of the mathematically correct ‘pi’ or the cuisine bound ‘pie’ the density will be light. In fact, in the case of ‘pi’ as it is, in fact, a number it holds no density whatsoever except in the case of a pre-pubescent schoolboy aching to attempt some simple maths, but instead watching a Sunday morning Open University special on integration and differentiation.

Suffice it to say that in the intervening period between this and my previous speculation on this matter, I have determined that the diameter of the Ryder is less than the distance from Wellington to Palmerston North as it took my clapped out Lada less than half the time to circumnavigate the Ryder than i?t took t??o get up the Kapiti Coast.

Mind you, I did st?op for an Ice Cream in Paekakariki.

Thank you for your erudite, illuminating and educating comme??nt, but you are overlooking one key factor.

Non-Newtonian phy?sics ?does allow for exceptional cases.

‘pi’ should not have been used in your formula – in the case of the Ryder it should have been ‘pie’ – and if it’s anything like the pies that Mrs Luckyeddie bakes, the crust is very, very dense indeed.

The average constituent component of the Ryder tends to be beer. Whilst beer is generally of a slightly higher density than water, on average, the Ryder consumes a significant amount of Heineken. The make up of Heineken is 50% piss, 30% C??arbon Dioxide (in free association with the water content of piss, making Carbonic Acid) and 20% Perrier. In fact, the density of Heineken is considerably less than water.

When an object the size of the Ryder is filled with Heineken (lets argue for the sake of simplicity th??at the Ryder is a sphere filled with pure Heineken) then the?? estimated volume of the Ryder is:

4/3 times Pi * radius cubed.

This is approximately 4.19 * the ?radius of the Ry??der cubed.

Given that density = mass / volume and that the volume of ??the Ryder is 4.19r^3 then we can deduce:

mass = density * 4.19r^3

Assuming?? that the density of water is 999 kg/m^3 and that Heineken is 30% Carbon Dioxide. (Ca??rbon Dioxide density = .00198 kg/m^3) then we can estimate the density of Heineken to be approximately 699 kg/m^3

Using this to solve the above equation (the Ryder Paradox), we can deduce that in order for Jesse Ryder to “create{s} a momentary indentation in the space-time continuum” then he’d have to be very, very big.

??In fact, if we assumed that his mass was 100?,000kg then at the estimated density, his radius would have to be

100000 = 699 * 4.19r^3

100000/699 = 4.19 * r^3

r = cube root (100000/699/4.19)

r = 3.24m

Hence, the diameter of the Ryder’s belly would need to be 6.48m to be even 100,000kg.

Given that at this size, his average temperature would be significantly raised by his own internal body heat, thus further reducing the average Heineken density, in practise this diameter w?ould be considerably greater.

I\’m not sure on the exact mass required to “create{s} a momentary indentation in the space-time continuum”, but I have a feeling it would be several factors larger than 100,000kg.

Hence, you are talking tripe.

😀 😀