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FFWD F3D FCC max tire width?

Internal rim width 22.4mm. (my mistake, 16mm in real)

In tech specs max recommended tire width is 35.

How do you think it`s ok to put 40c on this rims? I think 22.4mm is enough. or not?

If you're new please join in and if you have questions pop them below and the forum regulars will answer as best we can.

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10 comments

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BehindTheBikesheds | 6 years ago
1 like

I've just fitted a '40mm' tyre to a 17mm internal rim, it's coming out at 37mm at 30PSI (it's a wafer thin tyre probably would come up short on most rims), I've fitted 32mm to an open 4CD which are pretty narrow (12 or 13mm) and used it loaded and caning the 120degreee right hander at the mini roundabout a few hundred yards from my abode. Even with my 100kg I've never had issues with squirm etc.

EDIT, your 22.4mm is the EXTERNAL width, should still be okay though.

Avatar
lavds replied to BehindTheBikesheds | 6 years ago
0 likes

BehindTheBikesheds wrote:

I've just fitted a '40mm' tyre to a 17mm internal rim, it's coming out at 37mm at 30PSI (it's a wafer thin tyre probably would come up short on most rims), I've fitted 32mm to an open 4CD which are pretty narrow (12 or 13mm) and used it loaded and caning the 120degreee right hander at the mini roundabout a few hundred yards from my abode. Even with my 100kg I've never had issues with squirm etc.

EDIT, your 22.4mm is the EXTERNAL width, should still be okay though.

 

 

ohhh, my fault. 16mm internal, and now I'm more frightened by this fact)

Avatar
Acm | 6 years ago
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According to ETRTO, a 40mm tyre on a 22.4 run should work nicely.

However, wider tyres increase the stress (force for a given area) on the rim if used at the same pressure as a narrower tyre, which might be a problem on carbon rims.

I've never tested any of this in the real world and so don't know the numbers, this is all according to theory. I can't imagine that there would be a problem at low pressures (eg. cyclo-cross) but running them at 90psi could create issues (in theory).

Avatar
fukawitribe replied to Acm | 6 years ago
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Acm wrote:

However, wider tyres increase the stress (force for a given area) on the rim if used at the same pressure as a narrower tyre

Quick question - if the pressure is the same then in a uniform enclosure (e.g. a balloon) the stress would also be the same by definition wouldn't it ? What makes the difference in the (single) rim case - bead angle effects / effective cross-sectional tyre radius or something similar ?

Avatar
Canyon48 replied to fukawitribe | 6 years ago
1 like

fukawitribe wrote:

Acm wrote:

However, wider tyres increase the stress (force for a given area) on the rim if used at the same pressure as a narrower tyre

Quick question - if the pressure is the same then in a uniform enclosure (e.g. a balloon) the stress would also be the same by definition wouldn't it ? What makes the difference in the (single) rim case - bead angle effects / effective cross-sectional tyre radius or something similar ?

OoooOOOooo I think I might be able to answer this!

For any given pressure, the stress will increase as diameter increases (assuming isobaric and all of factors remain constant).

 

You can tell I'm fun at parties...   3

Avatar
fukawitribe replied to Canyon48 | 6 years ago
0 likes

wellsprop wrote:

fukawitribe wrote:

Acm wrote:

However, wider tyres increase the stress (force for a given area) on the rim if used at the same pressure as a narrower tyre

Quick question - if the pressure is the same then in a uniform enclosure (e.g. a balloon) the stress would also be the same by definition wouldn't it ? What makes the difference in the (single) rim case - bead angle effects / effective cross-sectional tyre radius or something similar ?

OoooOOOooo I think I might be able to answer this!

For any given pressure, the stress will increase as diameter increases (assuming isobaric and all of factors remain constant).

 

You can tell I'm fun at parties...   3

Haha - excellent ! ... although perhaps oddly I would find that fun at a party  4

That's a new concept for me, just been looking it up - that's the circumferential stress though isn't it, e.g. the tangential stress in the inner-tube, not the radial stress due to the pressure in the tyre ? Just trying to sketch it out in my head what the two effects mean for the static rim + beaded tyre case.... might be a while..

Avatar
Canyon48 replied to fukawitribe | 6 years ago
1 like

fukawitribe wrote:

wellsprop wrote:

fukawitribe wrote:

Acm wrote:

However, wider tyres increase the stress (force for a given area) on the rim if used at the same pressure as a narrower tyre

Quick question - if the pressure is the same then in a uniform enclosure (e.g. a balloon) the stress would also be the same by definition wouldn't it ? What makes the difference in the (single) rim case - bead angle effects / effective cross-sectional tyre radius or something similar ?

OoooOOOooo I think I might be able to answer this!

For any given pressure, the stress will increase as diameter increases (assuming isobaric and all of factors remain constant).

 

You can tell I'm fun at parties...   3

Haha - excellent ! ... although perhaps oddly I would find that fun at a party  4

That's a new concept for me, just been looking it up - that's the circumferential stress though isn't it, e.g. the tangential stress in the inner-tube, not the radial stress due to the pressure in the tyre ? Just trying to sketch it out in my head what the two effects mean for the static rim + beaded tyre case.... might be a while..

You're right, that is circumfrential stress not stress on the sidewall.

From wiki "The radial stress for a thick-walled cylinder is equal and opposite to the gauge pressure on the inside surface". Trouble is, a tyre/inner tube combination isn't a thick wall.

Thinking about it, I think you are correct about the force remaining the same regardless of tyre width for any pressure assuming the tyre/bead is parallel to the sidewall so, therefore, the force is normal.

This means the equation I made reference to is irrelevant and only serves purpose as material for mental masturbation  10

Avatar
fukawitribe replied to Canyon48 | 6 years ago
0 likes

wellsprop wrote:

fukawitribe wrote:

wellsprop wrote:

fukawitribe wrote:

Acm wrote:

However, wider tyres increase the stress (force for a given area) on the rim if used at the same pressure as a narrower tyre

Quick question - if the pressure is the same then in a uniform enclosure (e.g. a balloon) the stress would also be the same by definition wouldn't it ? What makes the difference in the (single) rim case - bead angle effects / effective cross-sectional tyre radius or something similar ?

OoooOOOooo I think I might be able to answer this!

For any given pressure, the stress will increase as diameter increases (assuming isobaric and all of factors remain constant).

 

You can tell I'm fun at parties...   3

Haha - excellent ! ... although perhaps oddly I would find that fun at a party  4

That's a new concept for me, just been looking it up - that's the circumferential stress though isn't it, e.g. the tangential stress in the inner-tube, not the radial stress due to the pressure in the tyre ? Just trying to sketch it out in my head what the two effects mean for the static rim + beaded tyre case.... might be a while..

You're right, that is circumfrential stress not stress on the sidewall.

From wiki "The radial stress for a thick-walled cylinder is equal and opposite to the gauge pressure on the inside surface". Trouble is, a tyre/inner tube combination isn't a thick wall.

Thinking about it, I think you are correct about the force remaining the same regardless of tyre width for any pressure assuming the tyre/bead is parallel to the sidewall so, therefore, the force is normal.

Aye maybe - or it might be that the bead is pulling more on the bead hook or something... or not, I keep flip-flopping on that...

 

wellsprop wrote:

This means the equation I made reference to is irrelevant and only serves purpose as material for mental masturbation  10

 1        Fair shout, it's still a jolly nice equation though !

Avatar
Canyon48 replied to fukawitribe | 6 years ago
0 likes
fukawitribe wrote:

wellsprop wrote:

=fukawitribe wrote:

wellsprop wrote:

fukawitribe wrote:

Acm wrote:

However, wider tyres increase the stress (force for a given area) on the rim if used at the same pressure as a narrower tyre

Quick question - if the pressure is the same then in a uniform enclosure (e.g. a balloon) the stress would also be the same by definition wouldn't it ? What makes the difference in the (single) rim case - bead angle effects / effective cross-sectional tyre radius or something similar ?

OoooOOOooo I think I might be able to answer this!

For any given pressure, the stress will increase as diameter increases (assuming isobaric and all of factors remain constant).

 

You can tell I'm fun at parties...   3

Haha - excellent ! ... although perhaps oddly I would find that fun at a party  4

That's a new concept for me, just been looking it up - that's the circumferential stress though isn't it, e.g. the tangential stress in the inner-tube, not the radial stress due to the pressure in the tyre ? Just trying to sketch it out in my head what the two effects mean for the static rim + beaded tyre case.... might be a while..

You're right, that is circumfrential stress not stress on the sidewall.

From wiki "The radial stress for a thick-walled cylinder is equal and opposite to the gauge pressure on the inside surface". Trouble is, a tyre/inner tube combination isn't a thick wall.

Thinking about it, I think you are correct about the force remaining the same regardless of tyre width for any pressure assuming the tyre/bead is parallel to the sidewall so, therefore, the force is normal.

Aye maybe - or it might be that the bead is pulling more on the bead hook or something... or not, I keep flip-flopping on that...

 

wellsprop wrote:

This means the equation I made reference to is irrelevant and only serves purpose as material for mental masturbation  10

 1        Fair shout, it's still a jolly nice equation though !

I reckon the bead "pulling" does exert a force in a real model, but I think it would be VERY difficult to create an accurate model. I imagine it will have a trig function of the angle the tyre makes with the sidewall.

Re: the hoop stress equation, the equation for longitudinal stress is Pd/4t, exactly half the hoop stress. This is why vessels always split lengthwise, rather than circumfrentially. Which is interesting (to me, anyway).

Avatar
fukawitribe replied to Canyon48 | 6 years ago
1 like

wellsprop wrote:

I reckon the bead "pulling" does exert a force in a real model, but I think it would be VERY difficult to create an accurate model. I imagine it will have a trig function of the angle the tyre makes with the sidewall.

 

I think you're right. I have an elegant proof for this, but it is too large to be contained within this thread.

 

wellsprop wrote:

Re: the hoop stress equation, the equation for longitudinal stress is Pd/4t, exactly half the hoop stress. This is why vessels always split lengthwise, rather than circumfrentially. Which is interesting (to me, anyway).

 

Damn, that is interesting. This is very cathartic you know...      4

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