Kite Skates

So are your axel-holes horizontal centered on the rail? Which would only leave 4x5 mm of steel above the axel hole, that does not sound like a lot, but if you say it works i do believe it does

Note im no expert in wielding and strength calculation of metals, so i could be way off here

Btw my skates is 5kg each

@JKS

For instance my rails are 30x8x560 mm long

//M

Gotchyar

For instance my rails are 30x8x560 mm long

//M

Yep big and heavy but strong.

Indeed spartan

Im working on a drawing of an aluminum frame, as far as my calculation goes the frame itself should be around 800 grams, then add wheels, bolts and boots. Should end up just around 2,5 kg. That would make a huge difference

How much does yours weight?

Edit:

Drawing in pdf can be downloaded here, all dimensions are in mm. Do you think 5mm aluminum is enough?

//M

Sorry M, Ive got no scales at home to weigh my skates.

I would go wider rather than thicker, eg 25mm wide x 3mm thickness,

cause there are huge down ward forces when jumping and landing.

Its easier to bend a 20mm x 5mm, but harder to bend a 25mm x 3mm.

Ok here is the stress side of things.

Let's assume the peak stress happens just next to the bolt on the rail and that it is a bending stress that is dominant, a reasonable assumption.

The maximum stress = 6*M*h/(t*(h^3-D^3)), p. 733 equ 7b, Roarks Formulas for Stress and Strain 6th Edition.

where
M = bending moment, N.mm. The bending moment here is proportional to your weight. So what works for George may not work for some one that is heavier.
h = height of the bar, mm
t = thickness of the bar, mm
D = hole diameter, mm
Stress = MPa units

So as George correctly pointed out, increasing h has the best effect to reduce the stress rather than thickness t.

So you could from this determine what sizes are needed if you use the same material that George used, provided George tells us his weight and you know how heavy you are. So all you do is plug in George's dimensions and weight (M) and out comes stress number (not a real stress because units of M is not mass, but it does not matter here). Then plug in your mass and dimensions and see if you exceed the stress from George's set up, if exceeding you need to alter your dimensions to come under his stress figures.

Regards,
Norman.

Ok here is the stress side of things.

Let's assume the peak stress happens just next to the bolt on the rail and that it is a bending stress that is dominant, a reasonable assumption.

The maximum stress = 6*M*h/(t*(h^3-D^3)), p. 733 equ 7b, Roarks Formulas for Stress and Strain 6th Edition.

where
M = bending moment, N.mm. The bending moment here is proportional to your weight. So what works for George may not work for some one that is heavier.
h = height of the bar, mm
t = thickness of the bar, mm
D = hole diameter, mm
Stress = MPa units

So as George correctly pointed out, increasing h has the best effect to reduce the stress rather than thickness t.

So you could from this determine what sizes are needed if you use the same material that George used, provided George tells us his weight and you know how heavy you are. So all you do is plug in George's dimensions and weight (M) and out comes stress number (not a real stress because units of M is not mass, but it does not matter here). Then plug in your mass and dimensions and see if you exceed the stress from George's set up, if exceeding you need to alter your dimensions to come under his stress figures.

Regards,
Norman.

Youre a real STAR Norman! That would make a great tool to see if im going in the right direction

//M

Sorry M, Ive got no scales at home to weigh my skates.

I would go wider rather than thicker, eg 25mm wide x 3mm thickness,

cause there are huge down ward forces when jumping and landing.

Its easier to bend a 20mm x 5mm, but harder to bend a 25mm x 3mm.

Yup was thinking the same - that also why im a little surprised that your rails that are only 20x5mm (Or are they 20x3 im a little unsure from your post at page 1?). As your reails leave only 4m "above" the axel-hole (Providing you use 12mm axel like me) im quite suprised that your rails dont deform

//M

Here's some close ups M,

but dont show these photos to anyone else, I'm keeping this system Top Secret...

PS that yellow thing under the boot is my kite peg, I cant forget it that way...

but dont show these photos to anyone else, I'm keeping this system Top Secret...

Oops ..... sorry George I peeked !




........ Working on memory loss as we speak

Here's some close ups M,

but dont show these photos to anyone else, I'm keeping this system Top Secret...

PS that yellow thing under the boot is my kite peg, I cant forget it that way...

Thank you - it seems youre rails is 5x20 mm and holds up quite well to your abuse

Clever one on the peg

//M

Just some further notes on the equations I gave:

To use the equations without modifying any factors, your system has to be the same as George's system for the following:
1) The number of rails used.
2) The distance from the wheel centre to where the first bolt to the skates is. Remember if you make that dimension longer, then the moment arm increases, increasing M, so that would need to be factored in as well if that dimension is changed.


Remember if you swap from Stainless steel to Aluminium there will be a strengh panalty. To give you an idea, the ratio of ultimate strength between good quality AL7050 and stainless steel can between 1/3.36 and 1/2.41, where as the ratio of density is 1/2.77. Also remember the AL at your typical hardware store will be rubbish and will not be as strong as AL7050.

Another stress check would be shear out at the end of the bar where the wheel bolt goes through, so using a similar by comparison approach the shear stress would be:

Shear stress = proportional to V/(t*(h-D))
V = shear load,which is proportional to rider weight.
t = bar thickness
h = bar height
D = hole diameter

Also remember you can not go crazy with making h too big without proper support, other wise the failure mode will be buckling instead of a stress failure.


Regards,
Norman.

Hey M, my skates weigh 3.9 kg each and the rail measures 520mm.

Now one other thing I need to explain.

I have got a second tyre and puncture liquid in each wheel to stop the punctures.

That would take the skates to 3.4 kg each.

Ok - ive been doin a bit of calculation on the rails

Assuming the point witch experiences the most stress is right where the front (Or back) mounting screw is - spartan's rails can withstand a vertically force of 107 kg and a horizontal force of 10 kg each. (Assuming he has not used any type of high grade steel).

To me that dosnt sound like much as he will put more force on them than his own weight when he lands a jump.

Well assuming the calculation is correct one could make a rail out of 6060-t6 aluminum with the following dimentions: 5x35mm - assuming there is no longer than 140 mm from the front (Or back) mounting screw to the axel

What do you think of that?

Edit: Well the force is on each point - assuming that the total force is distributed evenly that makes 4 points on each skate. Making it totally withstand above 400 kg

//M

This is a calculation only a Southerlybuster can work out.

Give me a problem to solve without numbers and I'll exel, throw in some numbers and I'll fail.

I dont have enough brain power in my left side of the brain to work out E=mc2.

Spartan.

It's not e=mc^2, it's s=My/I

Ok - ive been doin a bit of calculation on the rails

Assuming the point witch experiences the most stress is right where the front (Or back) mounting screw is - spartan's rails can withstand a vertically force of 107 kg and a horizontal force of 10 kg each. (Assuming he has not used any type of high grade steel).

To me that dosnt sound like much as he will put more force on them than his own weight when he lands a jump.

//M

What stress allowable did you use? A yield stress (when it yields) or ultimate (when it breaks)? I don't have the MMPDS at hand at the moment, but for steel (not stainless), elcheap grade is 125 MPa yield, 250 MPa for the better stuff and 350 MPa for the good stuff. From memory stainless steel is pretty strong and that is what George used.

Trying to guestimate the loads that your skates or vehicle will experience is always a challenge. Personally I go for what my body is likely to endure and anything beyond that there is no point making it stronger. So if you go for 2g loads, ie twice your body weight would be plenty (and some think this is over kill, and others think this is not enough!). That is the approach I used when designing my kiteboard. If you are serious about it, then put on some strain gauges. With the simple rail system that George uses, it's much easier to just estimate the load, make it, ride it and see if your assumptions were ok. If OK, then you can start optimising the design (ie do not exceed the peak stress) but reduce the weight, this then results in a fully stressed structure.

Regards,
Norman.

It's not e=mc^2, it's s=My/I

Norman I know, I'm mucking around

But what ever!!!... ....

Ok - ive been doin a bit of calculation on the rails

Assuming the point witch experiences the most stress is right where the front (Or back) mounting screw is - spartan's rails can withstand a vertically force of 107 kg and a horizontal force of 10 kg each. (Assuming he has not used any type of high grade steel).

To me that dosnt sound like much as he will put more force on them than his own weight when he lands a jump.

//M

What stress allowable did you use? A yield stress (when it yields) or ultimate (when it breaks)? I don't have the MMPDS at hand at the moment, but for steel (not stainless), elcheap grade is 125 MPa yield, 250 MPa for the better stuff and 350 MPa for the good stuff. From memory stainless steel is pretty strong and that is what George used.

Trying to guestimate the loads that your skates or vehicle will experience is always a challenge. Personally I go for what my body is likely to endure and anything beyond that there is no point making it stronger. So if you go for 2g loads, ie twice your body weight would be plenty (and some think this is over kill, and others think this is not enough!). That is the approach I used when designing my kiteboard. If you are serious about it, then put on some strain gauges. With the simple rail system that George uses, it's much easier to just estimate the load, make it, ride it and see if your assumptions were ok. If OK, then you can start optimising the design (ie do not exceed the peak stress) but reduce the weight, this then results in a fully stressed structure.

Regards,
Norman.

Well ive made the calculation using yield strength as i dont like the rails to be permanently deformed

What you need to know to calculate the dimensions:
- Yield strength (Y) of the material you want to use.
- Distance (D) from the outer part of the axel to the inner part of the first mounting hole (Mine is 140 mm).
- Thickness (t) of the material you want to use.
- Height (h) of the material you want to use.
- What "shape" the material is made in, eg L-profile, U-profil.

For a normal rail like spartans the following is true:
Max strength (kg) = (((Y*h^2*t)/D/2)*1/3)/9,8

For a rail made from L-profile the following is true:
Max strength (kg) = (((Y*h^2*t)/D/2)*0,63)/9,8

For instance - using a 3mm L-profile (35x35 mm) made from 6063-T6 aluminum the following would be true:
Y = 172 (According to wikipedia)
D = 140
t = 3
h = 35

Max strength = (((172*35^2*3)/140/2)*0,63)/9,8 = 145 kg pr rail.

Normal feel free to comment any

//M