
Re: Moment of Inertia of a Diamond/Double Cone
Jim Wilkinson Jul 21, 2014 7:50 AM (in response to Richie Sposato)Hi Richie,
I think that the PDF that you reference is incorrect, or unclear at best. I believe the formula should be Izaxis = 3/10 MR*2(R squared) for both cases of a cone and a double cone, as long as the mass being referenced is the mass of the cone in the first place and the double cone in the second. The reference PDF may be saying that the formula for a double cone is double that of the single cone because the mass doubles (so they are referencing the single cone mass in the second equation). But this is very misleading.I can't find another reference for the moment of inertia of a double cone, but there are many references for the moment of inertia of a sphere and of a half sphere (hemisphere) and the moment of inertia formulas for the Z direction are exactly the same for both. Here are two references that show this:
http://www.civilengineeringdegrees.net/2012/04/centerofgravitymassmomentof.html
http://engineeringreferences.sbainvent.com/dynamics/massmomentofinertia.php#.U8z7i7HD8eN
If a sphere and a half sphere have the same formula in Z, then the cone and double cone cases should as well since they are very similar types of geometry. Make sense?
I hope this helps,
Jim

Re: Moment of Inertia of a Diamond/Double Cone
Richie Sposato Jul 21, 2014 1:32 PM (in response to Jim Wilkinson)Hello Jim,
Thank you for taking the time to chime in on this.
I've spent many hours on this subject and as stated, there arent't any references for a double cone, which got me thinking that the formula used in solid works and other CAD programs is incorrect. One of the reasons is that the person who did the physics report actually did all the experiments by hand, so the numbers he came up with are true numbers, not numbers that came about from a program (that could be incorrect),
The reference to a sphere or a half sphere may sound similar to a double cone, but the difference is that the height of a double cone is independant from the width, whereas a sphere always has the same height as the width. If you increase the height of a sphere, you increase the radius. In a double cone, the height is independent of the radius. Therefore, when you increase the radius of the sphere, you're also increasing the height, which increases the mass linear, so the height cancels out.
This is why you can use the same formula for both a sphere and a 1/2 sphere, whereas you can't use it for a double cone.
Make sense, your thoughts..?
Richie

Re: Moment of Inertia of a Diamond/Double Cone
Jim Wilkinson Jul 21, 2014 2:57 PM (in response to Richie Sposato)Hi Richie,
I'm not so sure I agree with the argument of a sphere vs. a cone and height being a nonfactor for sphere, but not cone. If you look at the results from a cylinder, you'll also see that height has no impact on the moment of inertia about Z so if you chop a cylinder in half, the moment of inertia is only halved due to the mass being halved; the resulting moment is not divided in quarter due to the mass AND height each being divided in half. I think the same is true for this double cone case.I am not sure on the original PDF how the author determined on page 3 to multiply the result by 2 due to symmetry (and there are no calculations explaining that). If I go back to the formulas on page 1, it seems to me that the derivation of r across z would be done twice (ending up in a multiplying factor of 2), but then when you substitute in rho (mass/volume), the volume is also multiplied by 2 because it is two cones, not one. So, the new 2 on the top and 2 on the bottom (due to the volume) also cancel one another out in addition to the height, so the formula is the same. Unfortunately, the links in the PDF to the experimental information is not there so I can't look at that.
I'm also checking with some of the mathematicians here just to double check my thinking on it.
Thanks,
Jim

Re: Moment of Inertia of a Diamond/Double Cone
Richie Sposato Jul 21, 2014 3:31 PM (in response to Richie Sposato)Jim,
Let's try and grasp it this way, comparing a double cone/diamond to a sphere and a cone (1/2 diamond) to a hemisphere (1/2 sphere). A sphere is considered the main/whole shape, where the formula originates from, thus you start with that formula and extrapolate the MOI for a hemishpere off the formula of the sphere. The MOI formula for a sphere is 2/5 x M x RR, as published. It's reasonable to use the same formula for the hemisphere. As the mass changes, being 1/2 that of a sphere, using the same density and radius, the MOI will be 1/2 also.
It also seems reasonable to be able to double the value of a hemisphere MOI formula and add them together, because it's half a sphere. But diong this: (2/5 x M x RR) + (2/5 x M x RR) = 4/5 x M x RR, which is invalid, because it's already been established that the true formula for a sphere (or double hemisphere) is 2/5 x M x RR. Therefore, the "true" formula for a hemisphere is 1/5 x M x RR, whereas if you double it, 2 hemispheres = 1 sphere, i.e. (1/5 x M x RR) + (1/5 x M x RR) = 2/5 x Mm x RR. As you can see, this is valid as well, but to simplify things, the same formula (2/5 x M x RR) can be used for the hemisphere and sphere.
The problem arises and where the formula gets confused, is that the cone is considered the full/main shape, not the double cone/diamond, as with the sphere and hemisphere. It's been established and/or published that the true formula for a single cone is 3/10 x M x RR. It's also noted that the same formula for a single cone and double cone/diamond should be able to be used to calculate the respective MOI, same as a sphere and hemisphere. The formula used for both of these is assumed to be 3/10 x M x RR by CAD designers, because that's the published formula for the cone. This is understandable, because that's the only formula published, with knowledge that both being able to use the same formula in calculating their respective MOI.
Having said that, it's reasonable to conclude that the formula for a double cone/diamond can and should be (3/10 x M x RR) + (3/10 x M x RR) = 3/5 x M x RR, based on the same scenario as that of the sphere and the "true" hemishpere formula of (1/5 x M x RR) + (1/5 x M x RR) = sphere 2/5 x M x RR.
Because physics starts with a single cone that has a published MOI formula as the main shape and not a double cone (full sphere), you're allowed to doulbe it's formula to create a double cone/diamond and/or a full shape (i.e. sphere), whereas, you can't double the published formula for the hemisphere with it being 2/5 x M x RR.
With that said, because the MOI formula for the double cone isn't published and/or isn't a standard shape, and that it can use the same formula as a single cone, as seen with a sphere, CAD programmers inadvertently used the formula (3/10 x M x RR) for the double cone, instead of (3/5 x M x RR). Therefore, I've determined the CAD data for Solid Works and other CAD programs on the market concerning the MOI of a double cone/diaimond is incorrect. In conclusion, the correct formula to be used for the double cone/diamond is 3/5 x M x RR.
Does this make sense, your thoughts..?
Richie

Re: Moment of Inertia of a Diamond/Double Cone
Jerry Steiger Jul 21, 2014 4:30 PM (in response to Richie Sposato)Richie,
I agree with Jim.
I believe you are incorrect when you say that the true formula for a hemisphere is 1/5xMxRxR; it is 2/5xM/2xRxR, where M is the mass of the sphere. Similarly, the true formula for a double cone is 3/10xMxRxR, where M is the mass of the double cone, and the MOI for each single cone is 3/10xM/2xRxR where, again, M is the mass of the double cone.
Jerry S.

Re: Moment of Inertia of a Diamond/Double Cone
Jim Wilkinson Jul 22, 2014 8:24 AM (in response to Jerry Steiger)Hi Richie,
Jerry has it correct here. In those first two paragraphs, there is an error in your logic/formulas. To explain what Jerry is saying further, while R is a constant between the hemisphere and the sphere, M is not constant. In the case of the hemisphere, M is half of what it is in the sphere. So, lets call the mass of the hemisphere m and the mass of the sphere M. In that case, M=2m or m=M/2 (the mass of the sphere is equal to two hemispheres). So, using your first formula in the second paragraph, it would be (2/5 x m x RR) + (2/5 x m x RR) = 4/5 x m x RR. Then substitute in m=M/2 and you get 4/5 x M/2 x RR which equals 2/5 x M X RR.
The same logic can be applied to the double cone.
I hope this helps,
Jim





Re: Moment of Inertia of a Diamond/Double Cone
Jim, it's been a while since we've had this discussion and I'd like to revisit our conversation , if you're still here..?